Solving problems involving maxima and minima

In this section we'll apply the calculus of curve sketching to solve problems in which the solution is the maximum or minimum of some relevant function. The solutions to these problems will generally have three phases:

Solving max/min problems

1. Construct a function of one variable (see step 2) that models the quantity you're trying to minimize or maximize.


2. Take any necessary steps to reduce the function to one of one variable by finding expression(s) for one variable in terms of another.


3. Take the first derivative of the function and set it equal to zero; solve for the independent variable.

Example 1

A wire of length $L$ is to be cut into two pieces and bent into squares. Where should the wire be cut in order to maximize the total area of the two squares.

Solution: Here's a picture of what's going on. A wire of length $L$ is cut somewhere, and each piece is bent into two squares, of side lengths $x/4$ and $(L-x)/4$:

The squares look like this. Their areas are written below.

The area function that we will try to maximize is just the sum of these:

$$A(x) = \frac{x^2}{16} + \frac{(L - x)^2}{16}$$

The derivative of the area with respect to x (where the wire is cut) is

$$A'(x) = \frac{x}{8} + \frac{-(L - x)}{8}$$

The terms have the same denominator, so we can add them and set them equal to zero to find the critical point — the derivative is linear, so we expect only one.

$$\frac{2x - L}{8} = 0$$

The critical point is

$$x = \frac{L}{2}$$

Example 2

Johannes Kepler (1571-1630) proposed this problem: What are the dimensions of the largest wine barrel (cylinder) that can be inscribed inside of a sphere of radius R?

Solution: Here is a sphere of radius $R$ with a cylinder inscribed in it. Notice that the only points of contact between the sphere and cylinder are along the circular top and bottom of the cylinder. The radius of the cylinder is $r$ and its half-height is $h/2$.

The volume of any cylinder is a function of its radius and height:

$$V_{cyl} = \pi r^2 h$$

Now the Pythagorean theorem for the right triangle drawn in the figure above can be rearranged to find $r^2$,

$$\left( \frac{h}{2} \right)^2 + r^2 = R^2 \; \rightarrow \; r^2 = R^2 - \left( \frac{h}{2} \right)^2$$

With that last relationship, the volume function can be rewritten in terms of only the height:

$$V(h) = \pi \left( R^2 - \frac{h^2}{4} \right) h$$

We can clean that up a bit before taking a derivative, multiplying through by $\pi$ and $h$:

$$V(h) = \pi R^2 h - \frac{\pi h^3}{4}$$

The derivative with respect to the height is then

$$V'(h) = \pi R^2 - \frac{3 \pi h^2}{4}$$

Now to find the critical points, we set the first derivative equal to zero,

$$\frac{3 \pi h^2}{4} - \pi R^2 = 0$$

This quadratic function should give us two critical points, but one would give a negative height, so the single relevant root gives

$$h = \sqrt{\frac{4 R^2}{3}}$$

Taking the square root gives a compact version of $h$, one of our main results:

Example 3

An orchard owner has a plot of land with 25 apple trees planted on it. Agricultural studies predict that for every new tree planted, the yield of each tree will decrease by 12 apples. With 25 trees, each tree now averages 400 apples. How many trees should the farmer plant or remove in order to maximize the number of apples produced?

Here's the basic picture. More trees take more nutrients from existing trees and reduce the yield. It may be that a better yield will result from removing trees, but let's see.

First we construct a yield function, with an independent variable, $n$, representing the number of additional trees planted:

$$Y(n) = (n \; trees)(n \; apples)$$

The function is

$$Y(n) = (25 + n)(400 - 12n)$$

Example 4

A box without a top is to be constructed by cutting square flaps out of the corners of a 50 cm × 50 cm piece of cardboard, and folded as shown below. Find the dimensions of such a box that has the greatest volume.

The box is to be cut and folded as shown here. The "glue flaps" hold the sides at right angles to the bottom and to one another.

Here's a more fully-constructed version of the box, which will have a square bottom.

The layout of the flat piece of cardboard looks like this. The x-by-x squares are not really part of the surface area of the box; they overlap with the sides. The bottom of the box has sides measuring 50 - 2x cm, where x is the length of the square glue flaps.

The volume of our box, as a function of x, its height, is

$$ \begin{align} V(x) &= L \times W \times H \\ &= (50 - 2x)(50 - 2x) \cdot x \end{align}$$

Expanding that gives us a cubic function for the volume

$$ \begin{align} V(x) &= (2500 - 200x + 4x^2) \cdot x \\ &= 4x^3 - 200x^2 + 2500x \end{align}$$

Taking the first derivative gives

$$V'(x) = 12x^2 - 400x + 2500$$

Example 5

You are standing on a beach when you spot a drowning person. You are 150 m down the beach and the person is 25 m out into the water. You can run 9 m/s on the sand and you can swim at a rate of 1.5 m/s. How far should you run down the beach before swimming, in order to minimize the time it takes to reach the swimmer?

Here's a picture of this situation. This is actually quite an important problem; it's even useful in the field of optics. Of course, your brain pretty much does this calculation automatically in an emergency, but it's still fun to find the optimum path.

We begin by refining the diagram. We'll run a distance 150-x m along the beach before swimming. the resulting right triangle gives us a nice function for the total distance.

The distance is

$$L(x) = 150 - x + \sqrt{x^2 + 25^2}$$

Now the definition of speed can be rearranged to find the time.

$$s = \frac{d}{t} \; \longrightarrow \; t = \frac{d}{s}$$

We find a time function by dividing the two distances by their respective speeds, 9 m/s in sand (that's actually pretty fast in sand) and 1.5 m/s in the water.

Example 6

A tin can is to be manufactured so that it has a volume of 1 Liter. The material for the top and bottom costs $0.02/cm² and the material for the sides costs $0.01/cm². Find the dimensions of the can that minimizes its cost.

Here's a picture of the situation. The volume of a cylinder is $V(r,h)=\pi r^2 h$

One liter is 1000 cm³, so our volume equation is

$$1000 \; cm^3 = \pi r^2 h$$

Now we write a cost function representing the total cost of manufacturing a can. It's a function of two variables, r and h.

$$C = 0.01(2 \pi r h) + 2(0.02)(\pi r^2)$$

Using the volume constraint, solve for h:

$$h = \frac{1000}{\pi r^2}$$

and replace h in the cost function with it to get a cost function of one variable:

$$C(r) = \frac{0.01(2 \pi r)(1000)}{\pi r^2} = 0.04 \pi r^2$$

Reducing the function, we get

$$C(r) = \frac{20}{r} + 0.04 \pi r^2$$

Now take the derivative of that function with respect to r:

$$C'(r) = -\frac{20}{r^2} + 0.08 \pi r$$