Applied optimisation

Optimisation is the chapter's payoff. Real problems , minimising the surface area of a box, maximising the revenue of a firm , become exercises in calculus once we write the quantity to optimise as a function of a single variable. This subtopic walks through the recipe and a gallery of classical examples.

The recipe

1. Read carefully and identify what is to be maximised or minimised.
2. Draw a diagram if geometric.
3. Name the variables. Choose one as the independent variable.
4. Express the target quantity as a function of the chosen variable, using any constraints.
5. Differentiate and find critical points.
6. Classify critical points (1st or 2nd derivative test).
7. Check endpoints if applicable.
8. State the answer with units.

Classic example: largest rectangle in a semicircle

Find the largest rectangle that can be inscribed in a semicircle of radius $r$.

Let the rectangle have width $2x$ (so $-x \le X \le x$ where $X$ is the horizontal coordinate) and height $y$ where $x^2 + y^2 = r^2$.

Area $A = 2xy = 2x \sqrt{r^2 - x^2}$.

$\dfrac{dA}{dx} = 2\sqrt{r^2 - x^2} + 2x \cdot \dfrac{-x}{\sqrt{r^2 - x^2}} = \dfrac{2(r^2 - 2x^2)}{\sqrt{r^2 - x^2}}$.

Set to $0$: $r^2 = 2x^2$, $x = r/\sqrt 2$. Then $y = r/\sqrt 2$.

Maximum area: $2 \cdot r/\sqrt 2 \cdot r/\sqrt 2 = r^2$.

Box from a sheet

Make an open box from a square sheet of side $a$ by cutting squares of side $x$ from each corner and folding. Find the value of $x$ that maximises the volume.

After folding, the box has base $(a - 2x) \times (a - 2x)$ and height $x$.

$V(x) = x(a - 2x)^2$, $0 < x < a/2$.

$V'(x) = (a - 2x)^2 + x \cdot 2(a - 2x)(-2) = (a - 2x)(a - 2x - 4x) = (a - 2x)(a - 6x)$.

Set to $0$: $x = a/2$ (boundary, excluded) or $x = a/6$.

Maximum at $x = a/6$. Volume $= (a/6)(2a/3)^2 = 2a^3/27$.

Worked examples

Example 1. Find two positive numbers whose sum is $20$ and whose product is maximum.

$x + y = 20$, maximise $xy$. Substitute: $y = 20 - x$, $P = x(20 - x) = 20x - x^2$. $P' = 20 - 2x = 0 \Rightarrow x = 10$. So $x = y = 10$, max product $100$.

Example 2. Find the cylinder of largest volume inscribed in a sphere of radius $R$.

Cylinder: radius $r$, height $2h$. From sphere: $r^2 + h^2 = R^2$. Volume $V = \pi r^2 (2h) = 2\pi h (R^2 - h^2)$.

$V'(h) = 2\pi(R^2 - 3h^2) = 0 \Rightarrow h = R/\sqrt 3$, $r = R\sqrt{2/3}$.

Max volume: $\dfrac{4\pi R^3}{3\sqrt 3}$.

Example 3. A window has the shape of a rectangle topped by a semicircle. The perimeter is $10$ m. Find the dimensions that maximise the area of light admitted.

Let width $= 2r$, rectangular height $= h$. Perimeter: $2r + 2h + \pi r = 10$, so $h = (10 - 2r - \pi r)/2$.

Area $A = 2rh + \tfrac{\pi r^2}{2} = r(10 - 2r - \pi r) + \tfrac{\pi r^2}{2} = 10 r - 2 r^2 - \tfrac{\pi r^2}{2}$.

$\dfrac{dA}{dr} = 10 - 4r - \pi r = 0 \Rightarrow r = 10/(4 + \pi)$.

$h = (10 - 2 r - \pi r)/2 = 10/(4 + \pi)$. So $h = r$.

Width is $2r = 20/(4 + \pi)$.

Example 4. Find the point on $y = x^2$ closest to $(0, 1)$.

Distance squared: $D = x^2 + (x^2 - 1)^2$. Differentiate: $D' = 2x + 4 x (x^2 - 1) = 2x(1 + 2(x^2 - 1)) = 2x(2x^2 - 1)$.

Critical points: $x = 0$ or $x = \pm 1/\sqrt 2$. Values: $D(0) = 1$; $D(\pm 1/\sqrt 2) = 1/2 + 1/4 = 3/4$.

Closest points: $(\pm 1/\sqrt 2, 1/2)$.

Example 5. A wire of length $24$ m is cut into two pieces. One forms a square, the other a circle. Find the lengths to maximise total enclosed area.

Let one piece be $x$ (square side $x/4$) and the other $24 - x$ (circle radius $(24 - x)/(2\pi)$).

Area $A = (x/4)^2 + \pi \cdot ((24 - x)/(2\pi))^2 = x^2/16 + (24 - x)^2/(4\pi)$.

$dA/dx = x/8 - (24 - x)/(2\pi) = 0 \Rightarrow x = 96/(8 + \pi)$ at minimum. Check second derivative: $d^2 A/dx^2 = 1/8 + 1/(2\pi) > 0$. So this is a min.

Maximum is at an endpoint: $x = 0$ (all circle) gives $A = 24^2/(4\pi) = 144/\pi \approx 45.84$. $x = 24$ (all square) gives $A = 36$. So all wire on circle for max area.

Example 6. A manufacturer can sell $x$ items at $p = 5 - x/100$ rupees each. Cost is $C(x) = 500 + 2x$. Find $x$ that maximises profit.

Revenue $R(x) = x p = x(5 - x/100) = 5x - x^2/100$. Profit $P(x) = R - C = 3x - x^2/100 - 500$.

$P'(x) = 3 - x/50 = 0 \Rightarrow x = 150$. Profit: $3(150) - 150^2/100 - 500 = 450 - 225 - 500 = -275$.

Wait , negative profit means losing money. Let me re-examine: max profit = $-275$ means even at the optimal production level, the firm loses $275$ rupees. The firm should not produce.

(Or the problem could have intended different numbers; the procedure is correct.)

Try it yourself

1. Find two positive numbers whose product is $400$ and sum is minimum.
2. A rectangular box with a square base, open at the top, has volume $32$ cm$^3$. Find dimensions of minimum surface area.
3. Find the maximum area of an isosceles triangle inscribed in an ellipse with axes $a, b$.
4. A wire of length $L$ is bent into a rectangle. Find the side ratio that maximises area.
5. Find the shortest distance from $(0, c)$ to the parabola $y = x^2$.
6. A right circular cylinder is inscribed in a cone of height $h$ and base radius $r$. Find the cylinder's largest volume.
7. A man $1.8$ m tall walks toward a $4.5$-m lamp post at $1.2$ m/s. Find the rate at which the tip of his shadow moves.
8. A wire is bent to form a rectangle of perimeter $20$. Find the rectangle of maximum area.
9. Show: the rectangle of maximum perimeter that fits in a given square is the square itself.
10. A cone of height $h$ has slant length $\ell$ fixed. Find $h$ for maximum volume.
11. A man in a boat $3$ km from a straight shore wants to reach a point $5$ km up the shore in minimum time. He rows at $2$ km/h and walks at $5$ km/h. Where should he land?
12. Find the maximum area of a triangle with given perimeter.
13. Show that of all rectangles of given diagonal, the square has maximum area.
14. A box has square base and open top. Volume $108$ cm$^3$. Find dimensions of minimum surface area.

Pitfalls / Tricks

• Always write the target as a function of one variable using the constraint.
• Use the second derivative test or first derivative test to confirm max vs min.
• Check endpoints , the max often occurs at a boundary.
• Negative profit / negative cost / negative dimensions are nonsensical answers; check.
• For geometric problems, draw a picture and label all variables before computing.

This concludes the chapter. The next chapter on integrals is the antiderivative twin of differentiation.