Turning a stated problem into one function of one variable to optimise

What to assume they know

• Both derivative tests, and the closed-interval working rule
• Areas, surface areas and volumes of the rectangle, circle, cylinder, cone and sphere
• Similar triangles, and Pythagoras' theorem
• The equation of a parabola and of an ellipse in standard position
• Distance between two points in the plane
• Solving a quadratic, and discarding a root on physical grounds

What they should be able to do

• Identify, from a described problem, the quantity to be made largest or smallest
• Introduce a variable and express that quantity in terms of it and at most one other
• Find the relation in the description that ties the two unknowns together, and use it to eliminate one
• State the range of values the surviving variable may take, and justify discarding roots outside it
• Optimise a convenient related quantity — a square, or a square of a distance — instead of the one asked for, and say why that is legitimate
• Choose between the two derivative tests on the basis of the expression in front of you
• Draw and label a figure before writing any algebra, for problems the chapter supplies no figure for
• Answer in the terms the question asked — a dimension, a count, a ratio or a proof — rather than stopping at the critical value
• Recognise when the answer is required as a proof of a stated relation rather than as a number

Where it usually goes wrong

• "Start by differentiating." There is nothing to differentiate until one variable has been eliminated. Every worked example spends most of its length before the first derivative appears, and students who reach for the derivative first stall.
• "The constraint is extra information." It is the only thing that makes the problem solvable. Two unknowns and one expression cannot be optimised; the second relation is what reduces it to one.
• "Any critical value is the answer." Example 36 finds two and rejects one, because it would make a side of the box negative. The allowed range of the variable is part of the problem and the chapter mostly leaves it implicit.
• "I must optimise exactly what was asked for." Example 29 minimises a squared distance and Example 24's question is posed in squared distances from the start. Anything that only rises may be applied to both sides without moving the extremes — say why, then use it freely.
• "A negative answer just means I made a sign error." Sometimes it means the model does not fit. Exercise 6.3 Q6's profit function peaks at a negative count of units, and the honest response is to say so rather than to hide it.
• "The second test is always the right finisher." Example 25's second derivative runs half a page. When the first derivative is a quotient with a root in it, test its sign on each side instead.
• "Inscribed-solid problems each need their own trick." They share one: use the fixed quantity to eliminate, then optimise a square to clear the roots. Exercise 6.3 Q19 to Q26 are eight instances of that single pattern.
• "The answer is the value of the variable." Sometimes it is a ratio, a proof, a pair of numbers or a count. Read the last line of the question again before writing the last line of the answer.

Questions to check understanding

• Translate a described problem into a single function of a single variable, showing the eliminating relation
• State the range the variable may take, and reject a critical value that falls outside it
• Optimise a squared distance in place of a distance, and justify the substitution
• Solve a fixed-sum problem where the quantity involves unequal powers — the form of Exercise 6.3 Q14 and Q15
• Prove a stated relation for an inscribed solid — the form of Exercise 6.3 Q19 to Q26
• Maximise a profit function and say whether the answer is physically admissible
• Choose the nearest point of a printed curve to a stated point from four options — the form of Exercise 6.3 Q27

Examples worth working on the board

Values marked verified are worked out here from the chapter's own printed data; neither answers file was opened, and this chapter prints no answers to its own exercises.

• Example 22 (Part I pp. 167–168). Two positive numbers with a fixed sum of fifteen, minimising the sum of their squares. The whole translation is one substitution: call one number the variable and the other is the sum less it. Verified: the resulting function is twice the square, less thirty times the input, plus two hundred twenty-five; its derivative vanishes at seven and a half and its second derivative is the constant four, so both numbers are seven and a half. Section 4. This is the smallest possible instance of the pattern and the right place to name the four moves.
• Example 24 and Fig 6.16 (Part I p. 169). Two upright poles of stated heights at the ends of a stated span; a point on the span minimises the sum of the squared distances to the two tops. The constraint is that the two ground distances add to the span. Verified: with one written as the variable, the sum of squares is twice the square, less forty times the input, plus one thousand one hundred forty; the answer is ten, the midpoint. Section 7's first half: the question asks for a sum of squared distances, which is why no square root ever appears — a deliberate kindness.
• Example 25 and Fig 6.17 (Part I pp. 169–170). A trapezium with three sides of stated equal length, maximising the area. Two constructions are needed before any algebra: drop perpendiculars from the two ends of the shorter parallel side, and use the resulting congruence to see that the two outer feet cut off equal pieces. Verified: with that piece as the variable, the height follows from Pythagoras and the area is the variable plus ten, times the root of one hundred less its square; the critical value is five and the greatest area is seventy-five times the root of three. Section 8: the chapter supplies the figure, but a student meeting the problem cold would have to draw it, and the two dropped perpendiculars are the whole insight.
• Example 26 and Fig 6.18 (Part I pp. 170–171). A cylinder inscribed in a cone, maximising the curved surface. The eliminating relation is a similarity ratio between two triangles in the figure, which converts the cylinder's height into an expression in its radius. Verified: the surface becomes a constant times the cone's radius times the cylinder's radius, less the square of the cylinder's radius; the derivative vanishes when the cylinder's radius is half the cone's, and the second derivative is a negative constant. Section 5, and the model for every inscribed-solid item in the exercise.
• Example 29 (Part I p. 174). A point on a parabola nearest a stated fixed point. The chapter minimises the square of the distance rather than the distance, and notes there are no ends to add. Verified: the squared distance is the square of the horizontal displacement plus the fourth power of the input; its derivative factors with a single real root at one, and the value there is five, so the least distance is the square root of five. Section 7's second half: the square and the distance are largest and smallest at the same inputs because the square root only rises, and saying that out loud is worth more than the example.
• Example 36 and Fig 6.23 (Part I pp. 181–182). Equal corner squares removed from a rectangular sheet of stated dimensions, and the sides folded up to make an open box of greatest volume. The constraint is geometric and immediate: the base sides are the sheet's sides less twice the cut. Verified: the volume is the cut, times the shorter side less twice the cut, times the longer side less twice the cut; the derivative factors to give critical values of three and two thirds; three is rejected because it would make one base side negative, and at two thirds the second derivative is negative, so the greatest volume is two hundred over twenty-seven cubic metres. Fig 6.23 has two panels — the flat sheet with the four corner squares marked and the inner dimensions labelled, and the folded box in three dimensions with the same three labels. Section 6: the chapter rejects the larger root with a bracketed question rather than a reason.
• Example 37 (Part I pp. 182–183). A manufacturer whose unit price falls with the number sold, and whose cost is linear in it; the number maximising profit is wanted. Verified: revenue is five times the count less its square over a hundred, cost is a fifth of the count plus five hundred, so profit is twenty-four fifths of the count, less its square over a hundred, less five hundred; the derivative vanishes at two hundred forty and the second derivative is a negative constant. Section 9. The translation here is commercial rather than geometric, and the eliminating relation is that price depends on quantity — the only example in the chapter where the constraint is not a shape.
• Exercise 6.3 Q6 (Part I p. 175). A profit function printed as a downward quadratic in the number of units. Verified: the derivative vanishes at minus two and the value there is one hundred thirteen. The maximising input is negative, which no count of units can be; see Notes. Section 9 should use this as the counterpoint to Example 37: the same shape of problem, and an answer that fails the physical range the description implies.
• Exercise 6.3 Q13 to Q18 (Part I pp. 175–176). Six problems where the constraint is a stated sum. Verified: Q13, twelve and twelve. Q14, fifteen and forty-five, because the derivative of the product of one number and the cube of the other vanishes when the second is three quarters of the total. Q15, ten and twenty-five, by the same pattern with the exponents two and five splitting the total in the ratio two to five. Q16, eight and eight. Q17, a cut of three centimetres from a square sheet of side eighteen. Q18, a cut of five centimetres from a sheet of forty-five by twenty-four; the rejected root is eighteen, which would make one base side negative. Section 11 opens here, and Q14 and Q15 are worth putting side by side: the exponents decide the split.
• Exercise 6.3 Q19 to Q26 (Part I p. 176). Eight items where the answer is a stated relation to be proved rather than a number. Verified: Q19, the rectangle of greatest area in a fixed circle is the square. Q20, a cylinder of fixed total surface has greatest volume when its height equals its base diameter. Q21, the closed can of fixed volume with least surface has height twice its radius, and that radius is the cube root of fifty over pi. Q22, the wire is cut so the circle's piece is twenty-eight pi over pi plus four, and the square's piece is one hundred twelve over pi plus four. Q23, the greatest cone in a sphere has height four thirds of the radius and volume eight twenty-sevenths of the sphere's. Q24, the cone of least curved surface for a fixed volume has height the square root of two times its radius. Q25, the greatest-volume cone of fixed slant height has semi vertical angle the inverse tangent of the square root of two. Q26, for fixed surface area the corresponding angle is the inverse sine of one third. Section 11, and the shared pattern is worth stating: eliminate using the fixed quantity, then optimise the square of what you want.
• Exercise 6.3 Q27 (Part I p. 177). The nearest point of a printed parabola to a stated point, as four options. Verified: writing the squared distance in terms of the vertical coordinate gives a quadratic whose derivative vanishes at four, and the horizontal coordinate is then twice the square root of two — the first option. This is Example 29 with a different parabola and a different fixed point, and it is worth showing that the substitution which made it a quadratic is what saved the work.
• Miscellaneous Exercise Q5 to Q9 (Part I p. 184). Verified for Q5: with the apex sitting where the longer axis meets the curve, and the base drawn vertical, the triangle's area is the semi-minor axis over the semi-major, times the horizontal displacement, times the root of the difference of squares; it is greatest when the base sits at half the semi-major axis on the far side, and the area is three root three quarters of the product of the two semi-axes. Verified for Q6: the fixed depth and volume force a base area of four square metres, so the cost is a constant plus a multiple of the base perimeter, least when the base is square, and the total is one thousand rupees. Verified for Q7: writing the square's side from the fixed total perimeter gives a quadratic whose minimum puts the side at twice the circle's radius. Verified for Q8: with the rectangle's width twice the semicircle's radius and the perimeter fixed at ten, the area peaks when that radius is ten over four plus pi, and the rectangle's height equals it. Verified for Q9: minimising the hypotenuse through a point at stated distances from the two legs gives the stated three-halves power of the sum of two-thirds powers.
• Miscellaneous Exercise Q12, Q14 and Q15 (Part I p. 184). Three inscribed solids. Verified: Q12, the greatest-volume cone in a sphere has height four thirds of the radius — the same result Exercise 6.3 Q23 asks for, printed twice in one chapter. Q14, the greatest-volume cylinder in a sphere has height twice the radius over the square root of three, and volume four pi times the cube of the radius over three root three. Q15, the greatest-volume cylinder in a cone has height one third of the cone's, and the stated volume follows.
• The three problems in §6.4's opening (Part I p. 160). An orchard's profit, a thrown ball's greatest height, and a helicopter's nearest approach. Section 12 should return to them and be honest: the second is never worked anywhere in the chapter, and the third is worked with a different observer position and so reaches a different answer. The first is never worked either. See Notes.

Figures to have open

• A four-step card for section 2, to be recalled at each example. Not in the book.
• A redraw of Fig 6.16 (Part I p. 169): two upright poles of unequal height at the ends of a horizontal segment, the chosen point between them, and the two slant lines. The chapter's own.
• A redraw of Fig 6.17 (Part I p. 170): the trapezium with the two perpendiculars dropped, the right-angle marks kept, and the three equal sides and the two outer pieces labelled. The chapter's own, and it must be built up in stages for section 8.
• A redraw of Fig 6.18 (Part I p. 171): the cone with its inscribed cylinder, the axis dashed, and the two similar triangles highlighted. The chapter's own, with the highlighting added by the explanation.
• A redraw of Fig 6.23 (Part I p. 182), both panels: the flat sheet with corner squares and inner dimensions, and the folded open box with the same three labels. The chapter's own.
• Two curves for section 7, a distance and its square, with a shared vertical line through their common minimum. Not in the book.
• Sketches for Exercise 6.3 Q19 to Q26 — a rectangle in a circle, a cone in a sphere, a cylinder in a sphere, a cylinder in a cone. The chapter prints no figure for any of the eight, and a student cannot start without one.

Where this sits in the book

• NCERT Class 12 Mathematics, Part I, Chapter 6 "Application of Derivatives", §6.4's three opening problems, Part I p. 160
• Example 22, Part I pp. 167–168; Example 24 with Fig 6.16, Part I p. 169
• Example 25 with Fig 6.17, Part I pp. 169–170; Example 26 with Fig 6.18, Part I pp. 170–171
• Example 29, Part I p. 174
• Example 36 with Fig 6.23, and Example 37, Part I pp. 181–183
• Exercise 6.3, questions 6 and 13 to 27, Part I pp. 175–177
• Miscellaneous Exercise on Chapter 6, questions 5 to 9, 12, 14 and 15, Part I p. 184