Exercise 6.3 answers: Application of Derivatives

Open NCERT p. 174Checked by computer

(i) f(x) = (2x – 1)² + 3

1. f(x) = (2x − 1)² + 3 is a square plus 3, so it can never go below 3.
2. It equals 3 at x = 1/2, and grows without bound as x moves away from 1/2 on either side.
3. So there is a lowest value but no highest value, since the domain is all of R.

AnswerMinimum value 3 (at x = 1/2); no maximum value

(ii) f(x) = 9x² + 12x + 2

1. Write 9x² + 12x + 2 as a completed square: 9(x + 2/3)² − 2.
2. The squared term can never go below 0, so the expression can never go below −2.
3. It equals −2 at x = −2/3, and grows without bound elsewhere, with no upper limit on its domain R.

AnswerMinimum value −2 (at x = −2/3); no maximum value

(iii) f(x) = –(x – 1)² + 10

1. f(x) = −(x − 1)² + 10 is 10 minus a square, so it can never go above 10.
2. It equals 10 at x = 1, and decreases without bound as x moves away from 1.
3. So there is a highest value but no lowest value.

AnswerMaximum value 10 (at x = 1); no minimum value

(iv) g(x) = x³ + 1

1. g(x) = x³ + 1 has derivative g'(x) = 3x², which is zero only at x = 0.
2. But 3x² ≥ 0 on both sides of x = 0, so g is increasing throughout R with no turning point.
3. As x runs over all of R, g(x) runs over all of R too, with no highest or lowest value.

AnswerNo maximum value; no minimum value

Watch this explained “The square, and the end”, 7:25 into Local maxima and minima, and why critical points include the non-differentiable ones

Open NCERT p. 175Checked by computer

(i) f(x) = |x + 2| – 1

1. |x + 2| is never negative, and equals 0 when x = −2.
2. So f(x) = |x + 2| − 1 is never below −1, and equals −1 at x = −2.
3. As x moves away from −2 in either direction, f(x) grows without bound, so there is no highest value.

AnswerMinimum value −1 (at x = −2); no maximum value

(ii) g(x) = – |x + 1| + 3

1. |x + 1| is never negative, and equals 0 when x = −1.
2. So g(x) = −|x + 1| + 3 is never above 3, and equals 3 at x = −1.
3. As x moves away from −1, g(x) decreases without bound, so there is no lowest value.

AnswerMaximum value 3 (at x = −1); no minimum value

(iii) h(x) = sin(2x) + 5

1. sin(2x) always lies between −1 and 1, so h(x) = sin(2x) + 5 always lies between 4 and 6.
2. sin(2x) = 1 at x = π/4, where h = 6, and sin(2x) = −1 at x = 3π/4, where h = 4.

AnswerMaximum value 6; minimum value 4

(iv) f(x) = |sin 4x + 3|

1. sin 4x always lies between −1 and 1, so sin 4x + 3 always lies between 2 and 4 and is always positive.
2. So |sin 4x + 3| = sin 4x + 3, whose values run from 2 to 4.
3. sin 4x = 1 at x = π/8, giving 4, and sin 4x = −1 at x = 3π/8, giving 2.

AnswerMaximum value 4; minimum value 2

(v) h(x) = x + 1, x ∈ (– 1, 1)

1. h(x) = x + 1 rises steadily, so as x runs over (−1, 1) its values fill (0, 2).
2. The value 2 would need x = 1 and the value 0 would need x = −1, but neither end belongs to the open interval (−1, 1).
3. For any x in (−1, 1) there is a point further right inside the interval with a larger value, and a point further left with a smaller value, so no value is the greatest or the least.

AnswerNo maximum value; no minimum value

Watch this explained “A lowest value with no derivative”, 9:11 into Local maxima and minima, and why critical points include the non-differentiable ones

Open NCERT p. 175Checked by computer

(i) f(x) = x²

1. f'(x) = 2x, zero only at x = 0.
2. f''(x) = 2, which is positive, so x = 0 is a local minimum.
3. Local minimum value: f(0) = 0.

AnswerLocal minimum at x = 0, value 0; no local maximum

(ii) g(x) = x³ – 3x

1. g'(x) = 3x² − 3 = 3(x − 1)(x + 1), zero at x = 1 and x = −1.
2. g''(x) = 6x. At x = −1, g'' = −6 (negative) so it is a local maximum. At x = 1, g'' = 6 (positive) so it is a local minimum.
3. Values: g(−1) = −1 + 3 = 2, g(1) = 1 − 3 = −2.

AnswerLocal maximum at x = −1, value 2; local minimum at x = 1, value −2

(iii) h(x) = sin x + cos x, 0 < x < π/2

1. h'(x) = cos x − sin x, zero when tan x = 1, i.e. x = π/4 (inside the given domain).
2. h''(x) = −sin x − cos x, which at x = π/4 is negative, so x = π/4 is a local maximum.
3. Value: h(π/4) = sin(π/4) + cos(π/4) = √2/2 + √2/2 = √2. There is no other critical point in (0, π/2), so no local minimum.

AnswerLocal maximum at x = π/4, value √2; no local minimum

(iv) f(x) = sin x – cos x, 0 < x < 2π

1. f'(x) = cos x + sin x, zero when tan x = −1, i.e. x = 3π/4 or x = 7π/4 in (0, 2π).
2. f''(x) = cos x − sin x. At x = 3π/4 this is negative (local maximum); at x = 7π/4 this is positive (local minimum).
3. Values: f(3π/4) = sin(3π/4) − cos(3π/4) = √2/2 + √2/2 = √2. f(7π/4) = sin(7π/4) − cos(7π/4) = −√2/2 − √2/2 = −√2.

AnswerLocal maximum at x = 3π/4, value √2; local minimum at x = 7π/4, value −√2

(v) f(x) = x³ – 6x² + 9x + 15

1. f'(x) = 3x² − 12x + 9 = 3(x − 1)(x − 3), zero at x = 1 and x = 3.
2. f''(x) = 6x − 12. At x = 1 this is negative (local maximum); at x = 3 this is positive (local minimum).
3. Values: f(1) = 1 − 6 + 9 + 15 = 19. f(3) = 27 − 54 + 27 + 15 = 15.

AnswerLocal maximum at x = 1, value 19; local minimum at x = 3, value 15

(vi) g(x) = x/2 + 2/x, x > 0

1. g'(x) = 1/2 − 2/x², zero when x² = 4, so x = 2 (only x = 2 is in the domain x > 0).
2. g''(x) = 4/x³, which is positive at x = 2, so it is a local minimum.
3. Value: g(2) = 1 + 1 = 2. There is no other critical point for x > 0, so no local maximum.

AnswerLocal minimum at x = 2, value 2; no local maximum

(vii) g(x) = 1/(x² + 2)

1. g(x) = 1/(x² + 2) is largest exactly when x² + 2 is smallest, which happens at x = 0.
2. So g has a local maximum at x = 0, with value 1/(0 + 2) = 1/2.
3. As |x| grows the denominator only grows, so there is no local minimum.

AnswerLocal maximum at x = 0, value 1/2; no local minimum

(viii) f(x) = x√(1 – x), 0 < x < 1

1. f'(x) = √(1 − x) − x/(2√(1 − x)) = (2 − 3x)/(2√(1 − x)), zero at x = 2/3 (inside 0 < x < 1).
2. The derivative is positive for x < 2/3 and negative for x > 2/3, so x = 2/3 is a local maximum.
3. Value: f(2/3) = (2/3)·√(1/3) = 2/(3√3) = 2√3/9. There is no other critical point in (0, 1), so no local minimum.

AnswerLocal maximum at x = 2/3, value 2√3/9; no local minimum

Watch this explained “The eight worked, with both answers”, 17:50 into A sign change either side of a critical point decides which kind it is

Open NCERT p. 175One way to think about it

(i) f(x) = eˣ

1. f'(x) = eˣ, which is positive for every real x, so it is never zero.
2. A local maximum or minimum needs the derivative to be zero (or not exist) there, and eˣ is defined and non-zero everywhere.
3. So f has no critical point at all, hence no maximum or minimum.

In shortf'(x) = eˣ ≠ 0 for all x, so f has no maxima or minima

(ii) g(x) = log x

1. g'(x) = 1/x, defined and positive for every x in the domain x > 0.
2. Since g'(x) is never zero on its domain, g has no critical point.

In shortg'(x) = 1/x ≠ 0 for all x > 0, so g has no maxima or minima

(iii) h(x) = x³ + x² + x + 1

1. h'(x) = 3x² + 2x + 1.
2. This is a quadratic in x with discriminant 2² − 4·3·1 = 4 − 12 = −8, which is negative, so it has no real roots.
3. Since the coefficient of x² is positive and there are no real roots, h'(x) is positive for every real x, so h has no critical point.

In shorth'(x) = 3x² + 2x + 1 has negative discriminant and positive leading coefficient, so it is always positive; h has no maxima or minima

Watch this explained “The two-case result, scored”, 15:17 into Local maxima and minima, and why critical points include the non-differentiable ones

Open NCERT p. 175Checked by computer

(i) f(x) = x³, x ∈ [– 2, 2]

1. f'(x) = 3x², which is zero only at x = 0, but the sign of f' does not change there, so x = 0 is not an extremum inside the interval.
2. Compare the two endpoints instead: f(−2) = −8 and f(2) = 8.
3. The largest of these is the absolute maximum, the smallest is the absolute minimum.

AnswerAbsolute maximum 8 (at x = 2); absolute minimum −8 (at x = −2)

(ii) f(x) = sin x + cos x, x ∈ [0, π]

1. f'(x) = cos x − sin x, zero when tan x = 1, giving x = π/4 inside [0, π].
2. Evaluate f at the critical point and the two ends: f(0) = 1, f(π/4) = √2, f(π) = −1.
3. The largest is √2, the smallest is −1.

AnswerAbsolute maximum √2 (at x = π/4); absolute minimum −1 (at x = π)

(iii) f(x) = 4x – (1/2)x², x ∈ [–2, 9/2]

1. f'(x) = 4 − x, zero at x = 4, which lies inside [−2, 9/2].
2. Evaluate f at x = −2, x = 4, and x = 9/2: f(−2) = −10, f(4) = 8, f(9/2) = 63/8 = 7.875.
3. The largest is 8, the smallest is −10.

AnswerAbsolute maximum 8 (at x = 4); absolute minimum −10 (at x = −2)

(iv) f(x) = (x – 1)² + 3, x ∈ [–3, 1]

1. f'(x) = 2(x − 1), zero at x = 1, which is the right-hand end of the interval [−3, 1] itself.
2. So the only candidates are the two ends: f(−3) = 16 + 3 = 19 and f(1) = 0 + 3 = 3.
3. The larger value is the absolute maximum, the smaller is the absolute minimum.

AnswerAbsolute maximum 19 (at x = −3); absolute minimum 3 (at x = 1)

Watch this explained “Worked, not read off”, 16:38 into Largest and smallest over a closed interval: candidates inside plus the two ends

Open NCERT p. 175Matches NCERT’s answer

1. The profit rule is p(x) = 41 − 72x − 18x². The question puts no limit on x, so we look for the greatest value of p(x) over all real x.
2. Differentiate: p′(x) = −72 − 36x. Setting p′(x) = 0 gives x = −2.
3. Differentiate again: p″(x) = −36, which is negative at every x. So x = −2 gives a maximum, and as it is the only critical point, it gives the greatest value of all.
4. Put x = −2 back into p(x): p(−2) = 41 − 72(−2) − 18(−2)² = 41 + 144 − 72 = 113.
5. A note on the model: x = −2 would be a negative number of units, which a real company cannot make. If x had to be 0 or more, p′(x) would be negative the whole way and the greatest profit would be p(0) = 41. The question as set puts no such limit on x, so the answer is 113.

Answer113

Watch this explained “The answer the description cannot use”, 16:30 into Turning a stated problem into one function of one variable to optimise

Open NCERT p. 175Matches NCERT’s answer

1. f(x) = 3x⁴ − 8x³ + 12x² − 48x + 25.
2. f′(x) = 12x³ − 24x² + 24x − 48 = 12(x − 2)(x² + 2).
3. x² + 2 is always positive, so the only critical point in [0, 3] is x = 2.
4. Compare f at the two ends and this critical point: f(0) = 25, f(2) = −39, f(3) = 16.
5. The largest value is 25 (at x = 0); the smallest is −39 (at x = 2).

AnswerMaximum value 25 (at x = 0); minimum value −39 (at x = 2).

Watch this explained “The method, in four steps”, 8:13 into Largest and smallest over a closed interval: candidates inside plus the two ends

Open NCERT p. 175Matches NCERT’s answer

1. sin 2x reaches its greatest possible value, 1, whenever 2x = π/2 + 2kπ, for a whole number k.
2. So x = π/4 + kπ.
3. Inside [0, 2π], this gives x = π/4 and x = 5π/4.
4. At both points, sin 2x = 1, the largest the sine function can ever be.

Answerx = π/4 and x = 5π/4

Watch this explained “Worked, not read off”, 16:38 into Largest and smallest over a closed interval: candidates inside plus the two ends

Open NCERT p. 175Matches NCERT’s answer

1. Let f(x) = sin x + cos x. It repeats every 2π, so its greatest value over all x is its greatest value on [0, 2π].
2. f′(x) = cos x − sin x. Setting f′(x) = 0 gives tan x = 1, so in [0, 2π], x = π/4 or x = 5π/4.
3. Compare f at these two points and at the ends: f(0) = 1, f(π/4) = √2/2 + √2/2 = √2, f(5π/4) = −√2/2 − √2/2 = −√2, f(2π) = 1.
4. The largest of these values is √2, reached at x = π/4. So the maximum value of sin x + cos x is √2.

Answer√2

Watch this explained “One word, and what it costs”, 15:51 into The quicker second-derivative test, and the case where it tells you nothing

Open NCERT p. 175Matches NCERT’s answer

1. f(x) = 2x³ − 24x + 107. f′(x) = 6x² − 24 = 6(x² − 4), zero at x = 2 and x = −2.
2. On [1, 3]: compare the ends and the critical point inside, x = 2. f(1) = 85, f(2) = 75, f(3) = 89.
3. The maximum on [1, 3] is 89, at x = 3.
4. On [−3, −1]: compare the ends and the critical point inside, x = −2. f(−3) = 125, f(−2) = 139, f(−1) = 129.
5. The maximum on [−3, −1] is 139, at x = −2.

AnswerMaximum on [1, 3] is 89 (at x = 3); maximum on [−3, −1] is 139 (at x = −2).

Watch this explained “The method, in four steps”, 8:13 into Largest and smallest over a closed interval: candidates inside plus the two ends

Open NCERT p. 176Matches NCERT’s answer

1. x = 1 lies strictly inside [0, 2], so if it is where the maximum occurs, the derivative there must be zero.
2. f(x) = x⁴ − 62x² + ax + 9, so f′(x) = 4x³ − 124x + a.
3. f′(1) = 4 − 124 + a = 0, which gives a = 120.
4. Check: with a = 120, f(0) = 9, f(1) = 68, f(2) = 17 — f(1) = 68 is indeed the largest, confirming a = 120.

Answera = 120

Watch this explained “What must happen strictly inside”, 5:51 into Largest and smallest over a closed interval: candidates inside plus the two ends

Open NCERT p. 176Matches NCERT’s answer

1. f(x) = x + sin 2x. f′(x) = 1 + 2cos 2x. Setting f′(x) = 0 gives cos 2x = −1/2.
2. In [0, 2π], this happens at x = π/3, 2π/3, 4π/3, 5π/3.
3. Compute f at these four points and at the two ends, x = 0 and x = 2π.
4. f(0) = 0, f(π/3) ≈ 1.91, f(2π/3) ≈ 1.23, f(4π/3) ≈ 5.05, f(5π/3) ≈ 4.37, f(2π) = 2π ≈ 6.28.
5. The largest value is 2π, at the end x = 2π; the smallest is 0, at the end x = 0 — neither is at a turning point.

AnswerMaximum value 2π (at x = 2π); minimum value 0 (at x = 0).

Watch this explained “Four candidates, both winners at the ends”, 9:14 into Largest and smallest over a closed interval: candidates inside plus the two ends

Open NCERT p. 176Matches NCERT’s answer

1. Let the two numbers be x and 24 − x.
2. Their product is P(x) = x(24 − x) = 24x − x².
3. P′(x) = 24 − 2x. Setting P′(x) = 0 gives x = 12.
4. P″(x) = −2, which is negative, so x = 12 gives the largest product.
5. The other number is 24 − 12 = 12.

Answer12 and 12

Watch this explained “The smallest problem of the kind”, 3:20 into Turning a stated problem into one function of one variable to optimise

Open NCERT p. 176Checked by computer

1. Since x + y = 60, y = 60 − x. Maximise f(x) = x(60 − x)³ for 0 < x < 60.
2. f′(x) = (60 − x)³ − 3x(60 − x)² = (60 − x)²(60 − 4x).
3. Since 60 − x ≠ 0 here, f′(x) = 0 gives 60 − 4x = 0, so x = 15.
4. f″(x) = −2(60 − x)(60 − 4x) − 4(60 − x)², so f″(15) = 0 − 4(45)² = −8100, which is negative. So x = 15 gives the maximum.
5. y = 60 − 15 = 45.

Answerx = 15, y = 45

Watch this explained “The exponents decide the split”, 13:11 into Turning a stated problem into one function of one variable to optimise

Open NCERT p. 176Checked by computer

1. Since x + y = 35, y = 35 − x. Maximise f(x) = x²(35 − x)⁵ for 0 < x < 35.
2. f′(x) = 2x(35 − x)⁵ − 5x²(35 − x)⁴ = x(35 − x)⁴(70 − 7x) = 7x(35 − x)⁴(10 − x).
3. For 0 < x < 35, both x and (35 − x)⁴ are positive, so f′(x) = 0 only at x = 10.
4. For 0 < x < 10, f′(x) > 0 (f is rising); for 10 < x < 35, f′(x) < 0 (f is falling). So x = 10 gives the maximum.
5. y = 35 − 10 = 25.

Answerx = 10, y = 25

Watch this explained “The exponents decide the split”, 13:11 into Turning a stated problem into one function of one variable to optimise

Open NCERT p. 176Matches NCERT’s answer

1. Let the two numbers be x and 16 − x.
2. The sum of cubes is S(x) = x³ + (16 − x)³.
3. S′(x) = 3x² − 3(16 − x)², which simplifies to 96(x − 8).
4. Setting S′(x) = 0 gives x = 8.
5. S″(x) = 6x + 6(16 − x) = 96 at x = 8, which is positive, so x = 8 gives the minimum.
6. The other number is 16 − 8 = 8.

Answer8 and 8

Watch this explained “The smallest problem of the kind”, 3:20 into Turning a stated problem into one function of one variable to optimise

Open NCERT p. 176Matches NCERT’s answer

1. Let x cm be the side of the square cut from each corner, where 0 < x < 9.
2. The box's base is (18 − 2x) cm by (18 − 2x) cm, and its height is x cm.
3. Volume V(x) = x(18 − 2x)².
4. V′(x) = (18 − 2x)² − 4x(18 − 2x) = (18 − 2x)(18 − 6x).
5. Setting V′(x) = 0 gives x = 9 (rejected: the base would have side 0, so there is no box) or x = 3.
6. V″(x) = −2(18 − 6x) − 6(18 − 2x), so V″(3) = 0 − 6(12) = −72, which is negative. So x = 3 gives the maximum volume.

Answer3 cm

Watch this explained “The range you are not given”, 7:04 into Turning a stated problem into one function of one variable to optimise

Open NCERT p. 176Matches NCERT’s answer

1. Let x cm be the side of the square cut from each corner, where 0 < x < 12.
2. The box's base is (45 − 2x) cm by (24 − 2x) cm, and its height is x cm.
3. Volume V(x) = x(45 − 2x)(24 − 2x) = 4x³ − 138x² + 1080x.
4. V′(x) = 12x² − 276x + 1080 = 12(x − 5)(x − 18).
5. Setting V′(x) = 0 gives x = 5 or x = 18. Reject x = 18: it is more than 12, and the shorter side of the base would be 24 − 36 = −12.
6. V″(x) = 24x − 276, so V″(5) = 120 − 276 = −156, which is negative. So x = 5 gives the maximum volume.

Answer5 cm

Watch this explained “The range you are not given”, 7:04 into Turning a stated problem into one function of one variable to optimise

Open NCERT p. 176One way to think about it

1. Let the fixed circle have radius R, so its diameter is 2R.
2. Any rectangle inscribed in the circle has its diagonal equal to the diameter: if its sides are x and y, then x² + y² = (2R)².
3. Write x = 2R sinθ and y = 2R cosθ, where θ is the angle the diagonal makes with a side.
4. The area is A(θ) = xy = 4R² sinθ cosθ = 2R² sin2θ.
5. A(θ) is largest when sin2θ = 1, that is, at θ = π/4.
6. At θ = π/4, x = 2R sin(π/4) = R√2 and y = 2R cos(π/4) = R√2, so x = y.
7. Since the two sides come out equal, the rectangle of greatest area inscribed in the circle is a square.

In shortThe rectangle of greatest area inscribed in a fixed circle is a square (side R√2, where R is the circle's radius).

Watch this explained “The pattern the inscribed solids share”, 17:35 into Turning a stated problem into one function of one variable to optimise

Open NCERT p. 176One way to think about it

1. Let the cylinder have radius r and height h. Its total surface area S = 2πr² + 2πrh is fixed.
2. So h = (S − 2πr²)/(2πr).
3. Volume V = πr²h = πr² · (S − 2πr²)/(2πr) = Sr/2 − πr³.
4. V′(r) = S/2 − 3πr². Setting V′(r) = 0 gives S = 6πr².
5. V″(r) = −6πr, which is negative, so this r gives the maximum volume.
6. Put S = 6πr² into S = 2πr² + 2πrh: 2πr² + 2πrh = 6πr², so 2πrh = 4πr², which gives h = 2r.
7. 2r is the diameter of the base, so at maximum volume the height equals the diameter of the base.

In shortAt maximum volume h = 2r, so the height equals the diameter of the base.

Watch this explained “The four moves, named once”, 1:24 into Turning a stated problem into one function of one variable to optimise

Open NCERT p. 176Matches NCERT’s answer

1. Let the can have radius r cm and height h cm, with volume πr²h = 100, so h = 100/(πr²).
2. Surface area S(r) = 2πr² + 2πrh = 2πr² + 200/r.
3. S′(r) = 4πr − 200/r². Setting S′(r) = 0 gives r³ = 50/π, so r = (50/π)^(1/3).
4. S″(r) = 4π + 400/r³, which is positive, so this r gives the minimum surface area.
5. h = 100/(πr²), which works out to h = 2r = 2(50/π)^(1/3).

Answerr = (50/π)^(1/3) cm, h = 2(50/π)^(1/3) cm

Watch this explained “The four moves, named once”, 1:24 into Turning a stated problem into one function of one variable to optimise

Open NCERT p. 176Matches NCERT’s answer

1. Let x m be the length bent into a square, so (28 − x) m is bent into a circle, where 0 < x < 28.
2. The square has side x/4, so its area is x²/16.
3. The circle has circumference (28 − x), so its radius is (28 − x)/(2π) and its area is π · ((28 − x)/(2π))² = (28 − x)²/(4π).
4. Combined area A(x) = x²/16 + (28 − x)²/(4π).
5. A′(x) = x/8 − (28 − x)/(2π). Setting A′(x) = 0: x/8 = (28 − x)/(2π). Multiply both sides by 8π: πx = 4(28 − x), so πx + 4x = 112 and x = 112/(π + 4).
6. A″(x) = 1/8 + 1/(2π), which is positive, so this x gives the minimum combined area.
7. The other piece is 28 − 112/(π + 4) = (28π + 112 − 112)/(π + 4) = 28π/(π + 4).

AnswerSquare piece 112/(π + 4) m; circle piece 28π/(π + 4) m.

Watch this explained “Where the eliminating relation comes from”, 2:27 into Turning a stated problem into one function of one variable to optimise

Open NCERT p. 176One way to think about it

1. Let the cone have height h and base radius r, inscribed in a sphere of radius R.
2. From the right-angled triangle formed inside the sphere, r² = h(2R − h), for 0 < h < 2R.
3. Volume V(h) = (1/3)πr²h = (1/3)πh²(2R − h).
4. V′(h) = (1/3)π(4Rh − 3h²). Setting V′(h) = 0 gives h = 4R/3 (h = 0 is rejected).
5. V″(h) = (1/3)π(4R − 6h), which at h = 4R/3 is negative, so this h gives the maximum volume.
6. At h = 4R/3, r² = (4R/3)(2R/3) = 8R²/9.
7. Maximum volume = (1/3)π(8R²/9)(4R/3) = 32πR³/81.
8. The sphere's volume is (4/3)πR³, and (32πR³/81) ÷ (4πR³/3) = 8/27.

In shortThe largest inscribed cone's volume is 8/27 of the sphere's volume, occurring at height h = 4R/3.

Watch this explained “The pattern the inscribed solids share”, 17:35 into Turning a stated problem into one function of one variable to optimise

Open NCERT p. 176One way to think about it

1. Let the cone have base radius r, height h and slant height l, with given volume V = (1/3)πr²h. So h = 3V/(πr²).
2. Curved surface C = πrl, where l² = r² + h². C is positive, so C is least exactly where C² is least; working with C² avoids the square root.
3. C² = π²r²(r² + h²) = π²r⁴ + π²r² · 9V²/(π²r⁴) = π²r⁴ + 9V²/r².
4. d(C²)/dr = 4π²r³ − 18V²/r³. Setting this to 0 gives 4π²r⁶ = 18V², that is, 9V² = 2π²r⁶.
5. d²(C²)/dr² = 12π²r² + 54V²/r⁴, which is positive, so this r gives the least curved surface.
6. From h = 3V/(πr²), h² = 9V²/(π²r⁴). Put 9V² = 2π²r⁶: h² = 2π²r⁶/(π²r⁴) = 2r².
7. So h = √2 · r: the altitude is √2 times the radius of the base.

In shortAt the least curved surface, h = √2 · r: the altitude is √2 times the radius of the base.

Watch this explained “Optimising something easier”, 9:11 into Turning a stated problem into one function of one variable to optimise

Open NCERT p. 176One way to think about it

1. Let the cone have fixed slant height l and semi-vertical angle α, with 0 < α < π/2. Its radius is r = l sin α and its height is h = l cos α.
2. Volume V(α) = (1/3)πr²h = (1/3)πl³ sin²α cos α.
3. V′(α) = (1/3)πl³(2 sin α cos²α − sin³α) = (1/3)πl³ sin α (2cos²α − sin²α).
4. Since sin α ≠ 0, V′(α) = 0 gives 2cos²α = sin²α, that is, tan²α = 2, so tan α = √2 (α is between 0 and π/2).
5. 2cos²α − sin²α = 2 − 3sin²α, which is positive while sin²α < 2/3 and negative after. So V′(α) changes from positive to negative at tan α = √2, and this α gives the maximum volume.
6. So the semi-vertical angle is tan⁻¹ √2.

In shortThe semi-vertical angle of the cone of maximum volume is tan⁻¹ √2.

Watch this explained “The four moves, named once”, 1:24 into Turning a stated problem into one function of one variable to optimise

Open NCERT p. 176One way to think about it

1. Let the cone have base radius r, slant height l, height h and semi-vertical angle α, so sin α = r/l. Its total surface area S = πr² + πrl is fixed.
2. So l = (S − πr²)/(πr).
3. Volume V = (1/3)πr²h, with h² = l² − r². V is positive, so V is greatest exactly where V² is greatest.
4. V² = (1/9)π²r⁴(l² − r²) = (1/9)π²r⁴[(S − πr²)²/(π²r²) − r²] = (1/9)r²[(S − πr²)² − π²r⁴] = (1/9)(S²r² − 2πSr⁴).
5. d(V²)/dr = (1/9)(2S²r − 8πSr³) = (2Sr/9)(S − 4πr²). Setting this to 0 gives S = 4πr².
6. d²(V²)/dr² = (1/9)(2S² − 24πSr²). At S = 4πr² this is (1/9)(2S² − 6S²) = −4S²/9, which is negative, so this r gives the maximum volume.
7. With S = 4πr²: πrl = S − πr² = 3πr², so l = 3r.
8. So sin α = r/l = 1/3, and the semi-vertical angle is sin⁻¹(1/3).

In shortThe semi-vertical angle of the cone of maximum volume is sin⁻¹(1/3).

Watch this explained “Optimising something easier”, 9:11 into Turning a stated problem into one function of one variable to optimise

Open NCERT p. 177Matches NCERT’s answer

1. Any point on the curve x² = 2y can be written as (x, x²/2).
2. The square of its distance from (0, 5) is D = x² + (x²/2 − 5)². Minimising D also minimises the actual distance, since a square root never changes which of two non-negative numbers is smaller.
3. Differentiate: dD/dx = 2x + 2(x²/2 − 5)·x = x³ − 8x.
4. Set dD/dx = 0: x(x² − 8) = 0, so x = 0 or x = ±2√2.
5. At x = 0, the point is (0, 0) and D = 25. At x = 2√2, the point is (2√2, 4) and D = 8 + 1 = 9.
6. 9 is smaller than 25, so (2√2, 4) is the nearer point.

Answer(A) (2√2, 4)

Watch this explained “No ends at all”, 15:05 into Largest and smallest over a closed interval: candidates inside plus the two ends

Open NCERT p. 177Matches NCERT’s answer

1. Let f(x) = (1 − x + x²)/(1 + x + x²).
2. Differentiating (quotient rule) gives f'(x) = 2(x² − 1)/(1 + x + x²)².
3. f'(x) = 0 when x² = 1, i.e. x = 1 or x = −1.
4. f(1) = 1/3 and f(−1) = 3, so 1/3 is the smaller candidate.
5. To be sure 1/3 really is the smallest value anywhere, rewrite: f(x) − 1/3 = 2(x − 1)² / [3(1 + x + x²)].
6. 1 + x + x² = 3/4 + (x + 1/2)², a positive number plus a square, so it is always positive. And (x − 1)² is never negative.
7. So f(x) − 1/3 can never be negative, i.e. f(x) ≥ 1/3 for every real x, with equality exactly at x = 1.

Answer(D) 1/3

Watch this explained “No ends at all”, 15:05 into Largest and smallest over a closed interval: candidates inside plus the two ends

Open NCERT p. 177Matches NCERT’s answer

1. Let g(x) = x(x − 1) + 1 = x² − x + 1, so the given expression is g(x)^(1/3).
2. The cube root is an increasing function — a bigger number always has a bigger cube root — so the given expression is largest at the same x where g(x) itself is largest.
3. g(x) = x² − x + 1 is an upward-opening parabola in x, so on the closed piece [0, 1] its largest value sits at one of the two ends, not in the middle.
4. g(0) = 1 and g(1) = 1, so the largest value of g on [0, 1] is 1 (the smallest, at x = 1/2, is 3/4).
5. So the maximum of the given expression is 1^(1/3) = 1.

Answer(C) 1

Watch this explained “The method, in four steps”, 8:13 into Largest and smallest over a closed interval: candidates inside plus the two ends