Short and Sweet Calculus

4.7 Optimization

In a great many practical problems, we want to find the least amount of time, least cost, greatest benefit, optimum size, etc. In such problems, which are called optimization problems, we look for the maximum or minimum value of a function and the particular value of the variable that gives such a value. In this section, we outline a strategy to systematically attack optimization problems by using differential calculus.

Example 4.15. Suppose we have a piece of cardboard with the dimensions 48 in \(\times\) 18 in. To carry our stuff, we are planning to make an open box from the cardboard by cutting out small squares of equal size from each corner and folding up the sides (Figure 4.15). If we want the box to have the largest volume, how long we should make the cut?

image

Solution

Let

\(x=\) length of each side of the square,

\(V=\) volume of the open box.
We need to find a relationship between \(V\) and \(x\). Because the dimensions of the box are \(48-2x\), \(18-2x\), and \(x\) (Figure 4.17), the volume of the box is \[V=(48-2x)(18-2x)x=4x^{3}-132x^{2}+864x.\]

Because the dimensions of the box must be nonnegative, we have \[\begin{cases} 0\leq x\\ 0\leq18-2x & \Rightarrow\\ 0\leq48-2x \end{cases}0\leq x\leq9.\] Because \(V\) is continuous on the interval \([0,9]\), its absolute maximum occurs either at an endpoint \(x=0\), \(x=9\), or at a critical point. To obtain critical points \[\frac{dV}{dx}=12x^{2}-264x+864=0\] \[\Rightarrow x^{2}-22x+72=(x-4)(x-18)=0\] \[\Rightarrow x=4,\text{ or }x=18.\] However, \(x=18\not\in[0,9]\). Therefore, the only critical point in the interval \([0,9]\) is \(x=4\), and the absolute maximum of \(V\) is one of the following numbers.

\(x\) \(0\) \(4\) \(9\)
\(V(x)\) \(0\) \(1600\) \(0\)

Therefore, the maximum volume is obtained when \(x=4\).

The graph of \(V(x)\) is shown in Figure 4.18.

Example 4.16. What is shortest distance from the point \((0,2)\) to the parabola \(8-x^{2}=2y\)?

Solution

The distance between the point \((0,2)\) and a point \((x,y)\) is \[D=\sqrt{(x-0)^{2}+(y-2)^{2}.}\] Because \((x,y)\) is on the graph of \(x^{2}+8=-4y\), then \[(x,y)=\left(x,-\frac{1}{4}(x^{2}+8)\right)\] and the distance \(D\) is \[\begin{aligned} D & =\sqrt{x^{2}+\left(\frac{1}{2}(8-x^{2})-2\right)^{2}}\\ & =\sqrt{x^{2}+\left(2-\frac{1}{2}x^{2}\right)^{2}}.\end{aligned}\] We notice that the \(x\) value that minimizes \(D\) also minimizes the expression inside the radical. So instead of minimizing \(D\), we can minimize \(D^{2}\) (that is, the expression inside the radical) to avoid differentiation of a square root. Therefore, the function \(f\) to be minimized is \[\begin{aligned} f(x) & =x^{2}+\left(2-\frac{1}{2}x^{2}\right)^{2}\\ & =x^{2}+4-2x^{2}+\frac{1}{4}x^{4}\\ & =\frac{1}{4}x^{4}-x^{2}+4.\end{aligned}\] Because \(f(x)\) is defined for all real \(x\), there are no endpoints of the domain to consider. The derivative of \(f\) \[f'(x)=x^{3}-2x=x(x^{2}-2)\] is zero when \[x=0,\pm\sqrt{2}.\] To applying the First Derivative Test (Theorem 4.3), we need to determine the sign of \(f’\)

\(x\) \(-\infty\) \(-\sqrt{2}\) \(0\) \(\sqrt{2}\) \(+\infty\)
sign of \(x+\sqrt{2}\) \(-\) \(0\) \(+\) \(+\) \(+\) \(+\) \(+\)
sign of \(x\) \(-\) \(-\) \(-\) \(0\) \(+\) \(+\) \(+\)
sign of \(x-\sqrt{2}\) \(-\) \(-\) \(-\) \(-\) \(-\) \(+\) \(+\)
\(\therefore\) sign of \(f'(x)\) \(-\) \(0\) \(+\) \(0\) \(-\) \(0\) \(+\)
Increasing/Decreasing \(f(x)\) \(\searrow\) \(\nearrow\) \(\searrow\) \(\nearrow\)

It follows from the First Derivative Test and the above table that \(x=0\) yields a local maximum, whereas \(x=\pm\sqrt{2}\) yield a minimum distance. \[\left.D\right|_{x=\pm\sqrt{2}}=\sqrt{2+(2-\frac{1}{2}2)^{2}}=\sqrt{3}.\]

Let’s see what will happen if we use the Second Derivative Test (Theorem 4.6) instead of the First Derivative Test \[f”(x)=x^{2}-2\] Because \(f”(0)=-2<0\), \(f\) has a local maximum at \(x=0\). Because \(f”(\pm\sqrt{2})=0\), the Second Derivative Test (Theorem 4.6) fails, and we cannot determine if \(f\) has a local maximum, a local minimum at \(x=\pm\sqrt{2}\).

Therefore, \(f(\pm\sqrt{2})=\sqrt{3}\) is the shortest distance from the point \((0,2)\) to the parabola \(8-x^{2}=2y\), and the closest points on the parabola to \((0,2)\) are \((\sqrt{2},3)\) and \((-\sqrt{2},3)\).

The graph of \(D=\sqrt{x^{2}+(2-x^{2}/2)^{2}}\) is illustrated in Figure 4.19.

Let’s summarize the strategy that we have applied to solve the examples of this section:

Strategy for Solving Optimization Problems

  1. Identify the dependent variable that needs to be maximized or minimized and every variable that plays a role in the problem. Assign appropriate letters to the variables and constants that remind you of their actual meaning.

  2. Write a primary equation that relates the variable which is to be optimized to the other variables.

  3. If the primary equation is a function of more than one independent variable, eliminate the extra variables and reduce the equation to one having a single independent variable. This may involve the use of secondary equations relating the independent variables of the primary equation. The discovery of the secondary equations is often facilitated by drawing a figure or two.

  4. Determine the domain of the independent variable. This domain may be smaller than the natural domain of the function because of limitations inherent to the problem.

  5. Test the critical points and endpoints of the domain. Use the First Derivative Test or the Second Derivative Test to classify the critical points.


[up][previous][table of contents][next]