Mathematics 265 Introduction to Calculus I

Study Guide :: Unit 5

Applications of the Derivative

Optimization Problems

As we have seen (see Definition 5.9), the extreme values of a function correspond to the optimum values represented by the function. The type of problem we solve in this section requires us to find the maximum or the minimum values—which may turn to be the best or the worse, the largest or the smallest, etc.—of a function. That is, we must find the optimum value for a given situation.

The key to solving optimization problems is establishing the quantity $f$ that we want to maximize or minimize, and then constructing a function $f(x)$ that depends on a single quantity, $x$. Thus, if we want to maximize or minimize $f$, we look for the maximum or minimum value of $f$ for certain range of $x$. There are basically three ways to find these values. We have already considered two of them: finding the intervals of increase and decrease (see Remark 5.4), and the closed interval method. The third method is known as the second derivative test.

Second Derivative Test

Let ${\displaystyle f^{″}}$ be continuous at $c$. Then,

• if ${\displaystyle f^{\prime }\left(c\right)=0}$ and ${\displaystyle f^{″}\left(c\right)>0}$, then $f$ has a local minimum at $c$.
• if ${\displaystyle f^{\prime }\left(c\right)=0}$ and ${\displaystyle f^{″}\left(c\right)<0}$, then $f$ has a local maximum at $c$.

The methods we discussed earlier are always conclusive; however, the second derivative test may not be conclusive if ${\displaystyle f^{″}\left(c\right)=0}$. For instance, for ${\displaystyle f\left(x\right)=x^3}$ and ${\displaystyle g\left(x\right)=x^4}$, it is true that ${\displaystyle f^{\prime }\left(0\right)=0}$ and ${\displaystyle g^{\prime }\left(0\right)=0}$, but it is also true that ${\displaystyle f^{″}\left(0\right)=0}$ and ${\displaystyle g^{″}\left(0\right)=0}$. In the case of the function $f$, there is no local extreme value at $0$, and in the case of $g$, there is a local minimum at ${\displaystyle 0}$. Hence, when ${\displaystyle f^{″}\left(c\right)=0}$, we must use one of the other methods to find the extreme values.

Remember that the closed interval method only applies to continuous functions on closed intervals.

The graph of the quadratic function ${\displaystyle f\left(x\right)=a{x}^2+bx+c}$ with ${\displaystyle a\ne 0}$ is a parabola; hence, it has either an absolute maximum or an absolute minimum. We can use the second derivative test to see that this statement is true:

${\displaystyle f^{\prime }\left(x\right)=2ax+b;}$

hence,

${\displaystyle f^{\prime }\left(x\right)=0\text{ if }x=-{\displaystyle \frac{b}{2a}}}$

and

${\displaystyle f^{″}\left(x\right)=2a\text{.}}$

So, at

${\displaystyle x=-{\displaystyle \frac{b}{2a}}\text{,}}$

the function $f$ has an absolute minimum, if ${\displaystyle a>0}$, or an absolute maximum, if ${\displaystyle a<0}$.

Example 5.51.

1. The function ${\displaystyle f\left(x\right)=3x^2-7x}$ has an absolute minimum, because ${\displaystyle a=3>0}$, and that point is at

   ${\displaystyle x=-{\displaystyle \frac{-7}{2\left(3\right)}}={\displaystyle \frac{7}{6}}\text{.}}$

   So, the absolute minimum value is

   ${\displaystyle f\left({\displaystyle \frac{7}{6}}\right)=-{\displaystyle \frac{49}{12}}\text{.}}$

2. The function ${\displaystyle g\left(x\right)=\left(x-2\right)\left(4-x\right)}$ has an absolute maximum. [Why?] This value occurs at

   ${\displaystyle x={\displaystyle \frac{-6}{2\left(-1\right)}}=3\text{,}}$

   and the absolute maximum value is ${\displaystyle g\left(3\right)=1}$.

Example 5.52. If we consider the function ${\displaystyle f\left(x\right)=3x^2-7x}$ restricted to the interval ${\displaystyle \left[2,8\right]}$, we see from Example 5.51 that the absolute minimum is outside this interval. Hence, by the closed interval method, the absolute extreme values are at the endpoints: ${\displaystyle f\left(2\right)=-2}$ is the absolute minimum value, and ${\displaystyle f\left(8\right)=136}$ is the absolute maximum value.

Optimization problems always have limitations—that is, constraints. These constraints lead to a constraint equation that we must often establish to obtain a function $f(x)$ of only one variable $x$.

On page 240 of the textbook, the author presents a suggested strategy for solving optimization problems. Read it carefully, before continuing with the next examples.

Example 5.53. Read Example 1 on pages 240-241 of the textbook. In this Example the question is, “What are the dimensions of the field that has the largest area?” This tells us that we want to maximize the area ${\displaystyle A}$ of the field.

The area of the field depends on the dimensions $x$ (width) and $y$ (length); that is,

${\displaystyle A\left(x,y\right)=xy\text{.}}$

Since we need a function of ${\displaystyle A}$ with only one variable, we use the constraint conditions; in this case, the fenced perimeter of the field must be (is limited to) ${\displaystyle 2400}$ ft. Hence, the constraint equation is

${\displaystyle 2x+y=2400\text{.}}$

To obtain a function ${\displaystyle A}$ of one variable, we solve for either $x$ or $y$ from the constraint equation, and we substitute it into the function ${\displaystyle A\left(x,y\right)}$. In this case, we chose $y$, and ${\displaystyle A\left(x\right)=2400x-2x^2}$. Observe that ${\displaystyle 0<x<1200}$. [Why?]

The function ${\displaystyle A\left(x\right)}$ is a quadratic function; it has a maximum because the coefficient ${\displaystyle a=-2}$ is negative.

Thus, the absolute maximum for the function ${\displaystyle A\left(x\right)=2400x-2x^2}$ is at

${\displaystyle x=-{\displaystyle \frac{2400}{2\left(-2\right)}}=600}$

which is within the limits of ${\displaystyle 0}$ and ${\displaystyle 1200}$.

Note that ${\displaystyle x=600}$ is one of the dimensions of the field that gives the maximum area; the other is ${\displaystyle y=2400-2\left(600\right)=1200}$.

This result answers the question: the dimensions are ${\displaystyle 600\times 1200}$ ft. The maximum area for these dimensions is ${\displaystyle A\left(600\right)=720 000}$ ft.

Compare this solution with the one given in your textbook.

We did not need to use the second derivative test, since we know where the maximum or minimum occurs in the case of quadratic functions.

Example 5.54. [See Exercise 3 on page 245 of the textbook.]

Find two positive numbers whose product is ${\displaystyle 100}$ and whose sum is a minimum.

We want to minimize the sum ${\displaystyle S}$ of two positive numbers, say $x$ and $y$; hence, ${\displaystyle S\left(x,y\right)=x+y}$ is to be minimized with the constraint that their product must be ${\displaystyle 100}$. Therefore, ${\displaystyle xy=100}$ is the constraint equation.

Solving for $y$ from this equation we find that

${\displaystyle y={\displaystyle \frac{100}{x}}\text{.}}$

Therefore,

${\displaystyle S\left(x\right)=x+{\displaystyle \frac{100}{x}}}$

is a function of one variable, and ${\displaystyle 0<x<100}$. [Why?]

We need to find the minimum of this function. By the second derivative test, we have

${\displaystyle S^{\prime }\left(x\right)=1-{\displaystyle \frac{100}{x^2}}=0\text{if}x=10}$

and

${\displaystyle S^{″}\left(10\right)=200∕{\left(10\right)}^3>0\text{,}}$

so ${\displaystyle S}$ has a minimum if ${\displaystyle x=10}$ (which is within its range) and ${\displaystyle y=10}$.

We cannot use the closed interval method, because $x$ is on the open interval ${\displaystyle \left(0\text{, }100\right)\text{.}}$

Example 5.55. [See Exercise 31(a) on page 246 of the textbook.]

A piece of wire ${\displaystyle 10}$ m long is cut into two pieces. One piece is bent into a square, and the other into a equilateral triangle. How should the wire be cut such that the total area is a maximum?

We want to maximize the total area ${\displaystyle A}$ of the square and the triangle. [Recall, from Example 2.27, that the height of an equilateral triangle of side $y$ is ${\displaystyle {\displaystyle \frac{\sqrt{3}}{2}}y}$.]

If $x$ is the side of the square and $y$ is the side of the triangle, then

${\displaystyle A\left(x,y\right)=x^2+{\displaystyle \frac{\sqrt{3}}{4}}{y}^2\text{.}}$

The perimeters of the square and the triangle must be equal to the length of the wire; hence, ${\displaystyle 4x+3y=10}$. Solving for $y$, we find that

${\displaystyle y={\displaystyle \frac{10-4x}{3}}\text{,}}$

and the area is

${\displaystyle A\left(x\right)=x^2+{\displaystyle \frac{\sqrt{3}{\left(10-4x\right)}^2}{36}}\text{.}}$

Doing the operations and simplifying, we find

${\displaystyle A\left(x\right)={\displaystyle \frac{9+4\sqrt{3}}{9}}{x}^2-{\displaystyle \frac{20\sqrt{3}}{9}}x+{\displaystyle \frac{25\sqrt{3}}{9}}\text{ for }0\le x\le 2\text{.}5\text{.}}$

The area function ${\displaystyle A}$ is a quadratic with an absolute minimum, and we are looking for the absolute maximum of ${\displaystyle A}$. Thus, we use the closed interval method on ${\displaystyle \left[0,10\right]}$. The maximum is at the endpoints, since the absolute minimum is at the critical point. However,

${\displaystyle A\left(0\right)={\displaystyle \frac{25\sqrt{3}}{9}}\approx 4\text{.}81\text{and}A\left(2\text{.}5\right)={\left(2\text{.}5\right)}^2+{\displaystyle \frac{\sqrt{3}{\left(10-4\left(2\text{.}5\right)\right)}^2}{36}}={\displaystyle \frac{25}{4}}\text{.}}$

Therefore, the maximum area occurs when ${\displaystyle x=2\text{.}5}$. This result means that the wire should be used to do only a square.

Note: Optimization problems may be challenging: you must identify your variables and use whatever you know that relates to the problem. Nothing is gained if you do not learn to overcome challenges. Remember that time and dimensions are positive variables.

Tutorial 1: Optimization

Tutorial 2: Optimization