Optimization Problems

One quantity, one variable

An optimization problem asks for the largest or smallest value of some quantity: the most area, the least material, the lowest cost. The quantity usually depends on two or more measurements, tied together by a condition, the constraint.

A rectangle is to be fenced with 40 meters of fence. The quantity is its area, A = xy, where x and y are the width and the length in meters. The constraint is the perimeter: 2x + 2y = 40, so x + y = 20 and y = 20 − x.

Now the area depends on x alone: $A=x\left(20-x\right)=20x-x^2$A = x(20 − x) = 20x − x². Whatever width is chosen, the length follows. The width must be positive and leave a positive length, so the domain is 0 < x < 20.

One highest point

$A=20x-x^2$A = 20x − x² is a parabola opening downwards. At x = 4 the area is 4 × 16 = 64 square meters; at x = 9 it is 99; near either end of the domain it shrinks toward 0, as the rectangle flattens into a line. Somewhere between, the curve has one highest point.

A rectangle of width x and length 20 − x, with its area $A=20x-x^2$A = 20x − x² plotted against x. Drag the width: the area is 0 at the two ends of the range, and largest, 100 square meters, at x = 10, where $\frac{dA}{dx}=20-2x$dA/dx = 20 − 2x is 0.

Differentiate, then justify

$\frac{dA}{dx}=20-2x$dA/dx = 20 − 2x, which is 0 when x = 10. Then the length is 20 − 10 = 10, so the rectangle is a square.

A stationary point is not yet a maximum. Here there are two ways to be sure. The second derivative, $\frac{d^{2}A}{d{x}^2}=-2$d²A/dx² = −2, is negative, so the curve bends downwards and x = 10 is a maximum. Or compare: at the ends of the domain the area falls to 0, and at x = 10 it is 100, and the only stationary point between them is x = 10.

Then answer the question that was asked. The largest area is 100 square meters, from a 10 m by 10 m square. x = 10 is the width, not the answer.

The method

Name the quantity to be made largest or smallest, and give each measurement a letter. Write the constraint, and use it to put the quantity in terms of one variable. State the domain of that variable, from what the measurements can physically be.

Differentiate, set the derivative to 0, and solve, keeping only solutions in the domain. Justify the kind of point: the second derivative, or the values at the ends of a closed domain. Then give the value asked for, with its units.

An open box

An open box has a square base of side x meters and height h meters, and must hold 32 cubic meters. Find the least material, its surface area S.

The quantity is $S=x^2+4xh$S = x² + 4xh: the base and four sides. The constraint is the volume, $x^{2}h=32$x²h = 32, so $h=\frac{32}{x^2}$h = 32/x², and $S=x^2+4x\times \frac{32}{x^2}=x^2+\frac{128}{x}$S = x² + 4x × 32/x² = x² + 128/x. The domain is x > 0, with no upper limit.

$\frac{dS}{dx}=2x-\frac{128}{x^2}$dS/dx = 2x − 128/x², which is 0 when $2x^3=128$2x³ = 128, so $x^3=64$x³ = 64 and x = 4. Then $h=\frac{32}{16}=2$h = 32/16 = 2.

The second derivative is $\frac{d^{2}S}{d{x}^2}=2+\frac{256}{x^3}$d²S/dx² = 2 + 256/x³, which is positive for every x > 0. So the curve bends upwards along its whole domain, and its one stationary point is its lowest point overall. The least material is $S=16+\frac{128}{4}=16+32=48$S = 16 + 128/4 = 16 + 32 = 48 square meters, for a base 4 m square and a height of 2 m. Other bases cost more: x = 2 needs 4 + 64 = 68, and x = 8 needs 64 + 16 = 80.

The gold curve $S=x^2+\frac{128}{x}$S = x² + 128/x for x > 0, with the vertical scale squeezed. It falls from very large values near x = 0 to its lowest point (4, 48) and rises after it, through (2, 68) and (8, 80).

When an end wins

A wire 20 cm long is cut in two. A piece x cm long is bent into a square, of side $\frac{x}{4}$x/4, and the rest, 20 − x cm, into a circle, of radius $\frac{20-x}{2\pi }$(20 − x)/(2π). Find the least and the greatest total area.

The total area is $A=\frac{x^2}{16}+\pi {\left(\frac{20-x}{2\pi }\right)}^2=\frac{x^2}{16}+\frac{{\left(20-x\right)}^2}{4\pi }$A = x²/16 + π((20 − x)/(2π))² = x²/16 + (20 − x)²/(4π) square centimeters. Here x = 0 (all circle) and x = 20 (all square) are allowed, so the domain is the closed interval [0, 20].

The gradient of A is $\frac{x}{8}-\frac{20-x}{2\pi }$x/8 − (20 − x)/(2π), and setting it to 0 gives $\pi x=4\left(20-x\right)$πx = 4(20 − x), so $x\left(\pi +4\right)=80$x(π + 4) = 80 and $x=\frac{80}{\pi +4}\approx 11.20$x = 80/(π + 4) ≈ 11.20. The second derivative, $\frac{1}{8}+\frac{1}{2\pi }\approx 0.284$1/8 + 1/(2π) ≈ 0.284, is positive, so this is a minimum: $A\approx 14.00$A ≈ 14.00 square centimeters.

The greatest area is not at the stationary point, so compare the ends. At x = 0, $A=\frac{400}{4\pi }=\frac{100}{\pi }\approx 31.83$A = 400/(4π) = 100/π ≈ 31.83; at x = 20, $A=\frac{400}{16}=25$A = 400/16 = 25. The greatest total area is about 31.83 square centimeters, with no cut at all: the whole wire made into a circle.

The gold curve $A=\frac{x^2}{16}+\frac{{\left(20-x\right)}^2}{4\pi }$A = x²/16 + (20 − x)²/(4π) from x = 0 to x = 20. Its stationary point is the lowest point, about (11.20, 14.00); the highest value is at the end x = 0.

The usual mistakes

Differentiating before using the constraint. A = xy has two letters; until y is written as 20 − x, there is no single derivative to set to 0.

Giving the variable instead of the quantity. For the fence, x = 10 is a width; the largest area is 100 square meters.

Taking a stationary point as the answer without checking it. For the wire, the stationary point gives the least area, and the question asked for the greatest as well.

Putting the whole half-perimeter into one side. A side of 20 leaves a length of 0, and the area is 0.

Leaving off the units, or giving square meters for a length.

A can of fixed capacity

In the application below, a closed can must hold $128\pi$128π cubic centimeters. The capacity is the constraint that puts the metal in terms of the radius alone.