4.5 Optimization Applications

Example

The manager of a garden store wants to build a 600 square foot rectangular enclosure on the store’s parking lot in order to display some equipment. Three sides of the enclosure will be built of redwood fencing, at a cost of $7 per running foot. The fourth side will be built of cement blocks, at a cost of $14 per running foot. Find the dimensions of the least costly such enclosure.

Max-Min Story Problem Technique

1. Translate the English statement of the problem line by line into a picture (if that applies) and into math. This is often the hardest step!
2. Identify the objective function. Look for words indicating a largest or smallest value.
   1. If you seem to have two or more variables, find the constraint equation. Think about the English meaning of the word constraint, and remember that the constraint equation will have an equals sign.
   2. Solve the constraint equation for one variable and substitute into the objective function. Now you have an equation of one variable.
3. Use calculus to find the optimum values. (Take derivative, find critical points, test. Don’t forget to check the endpoints!)
4. Look back at the question to make sure you answered what was asked. Translate your number answer back into English.

Example 1

The manager of a garden store wants to build a 600 square foot rectangular enclosure on the store’s parking lot in order to display some equipment. Three sides of the enclosure will be built of redwood fencing, at a cost of $7 per running foot. The fourth side will be built of cement blocks, at a cost of $14 per running foot. Find the dimensions of the least costly such enclosure.

First, translate line by line into math and a picture:

The objective function is the cost function, and we want to minimize it. As it stands, though, it has two variables, so we need to use the constraint equation. The constraint equation is the fixed area [latex]A = xy = 600[/latex]. Solve [latex]A[/latex] for [latex]x[/latex] to get [latex]x=\frac{600}{y}[/latex], and then substitute into [latex]C[/latex]: \[C=14\left(\frac{600}{y}\right)+21y=\frac{8400}{y}+21y.\]

Now we have a function of just one variable, so we can find the minimum using calculus.

\[C’=-\frac{8400}{y^2}+21\] [latex]C'[/latex] is undefined for [latex]y = 0[/latex], and [latex]C' = 0[/latex] when [latex]y = 20[/latex] or [latex]y = -20[/latex].

Of these three critical numbers, only [latex]y = 20[/latex] makes sense (is in the domain of the actual function) – remember that [latex]y[/latex] is a length, so it can’t be negative, and [latex]y = 0[/latex] would mean there was no enclosure at all, so it couldn’t have an area of 600 square feet.

Test [latex]y = 20[/latex] (here we chose the second derivative test): \[C”=\frac{16800}{y^3} \gt 0,\] so this is a local minimum.

Since this is the only critical point in the domain, this must be the global minimum. Going back to our constraint function, we can find that when [latex]y = 20[/latex], [latex]x = 30.[/latex] The dimensions of the enclosure that minimize the cost are 20 feet by 30 feet.

Example 2

A concert promoter has found that if she sells tickets for $50 each, she can sell 1200 tickets, but for each $5 she raises the price, 50 less people attend. What price should she sell the tickets at to maximize her revenue?

We are trying to maximize revenue, and we know that [latex]R=pq[/latex], where [latex]p[/latex] is the price per ticket, and [latex]q[/latex] is the quantity of tickets sold.

The problem provides information about the demand relationship between price and quantity – as price increases, demand decreases. We need to find a formula for this relationship. To investigate, let’s calculate what will happen to attendance if we raise the price:

You might recognize this as a linear relationship. We can find the slope for the relationship by using two points: \[m=\frac{1150-1200}{55-50}=\frac{-50}{5}=-10.\]

You may notice that the second step in that calculation corresponds directly to the statement of the problem: the attendance drops 50 people for every $5 the price increases.

Using the point-slope form of the line, we can write the equation relating price and quantity: \[q-1200=-10(p-50).\]

Simplifying to slope-intercept form gives the demand equation \[ q=1700-10p. \]

Substituting this into our revenue equation, we get an equation for revenue involving only one variable: \[R=pq=p(1700-10p)=1700p-10p^2.\]

Now, we can find the maximum of this function by finding critical numbers. [latex]R'=1700-20p[/latex], so [latex]R'=0[/latex] when [latex]p = 85[/latex].

Using the second derivative test, [latex]R''=-20 \lt 0[/latex] (for any value of [latex]p[/latex]), so the critical number is a local maximum. Since it is the only critical number, we can also conclude that it is the global maximum.

The promoter will be able to maximize revenue by charging $85 per ticket. At this price, she will sell [latex]q=1700-10(85)=850[/latex] tickets, generating $72,250 in revenue.

Example 3

Suppose we want to maximize profit.

Now we know what to do – find the profit function, find its critical points, test them, etc.

But remember that Profit = Revenue – Cost. So Profit’ = Revenue’ – Cost’. That is, the derivative of the profit function is [latex]MR - MC[/latex].

Now let’s find the critical points – those will be where Profit’ = 0 or is undefined. Profit’ = 0 when [latex]MR - MC = 0[/latex], or where [latex]MR = MC[/latex].

Profit has critical points when Marginal Revenue and Marginal Cost are equal.

Example 4

A company sells [latex]q[/latex] ribbon winders per year at [latex]\$p[/latex] per ribbon winder. The demand function for ribbon winders is given by: [latex]p=300-0.02q[/latex]. The ribbon winders cost $30 apiece to manufacture, plus there are fixed costs of $9000 per year. Find the quantity where profit is maximized.

We want to maximize profit, but there isn’t a formula for profit given. So let’s make one. We can find a function for Revenue = [latex]pq[/latex] using the demand function for [latex]p[/latex]. \[R(q)=(300-0.02q)q=300q-0.02q^2.\]

We can also find a function for Cost, using the variable cost of $30 per ribbon winder, plus the fixed cost: \[C(q)=9000+30q.\]

Putting them together, we get a function for Profit: \[P(q)=R(q)-C(q)=\left(300q-0.02q^2\right)-(9000+30q)=-0.02q^2+270q-9000\]

Now we have two choices. We can find the critical points of Profit by taking the derivative of [latex]P(q)[/latex] directly, or we can find [latex]MR[/latex] and [latex]MC[/latex] and set them equal. (Naturally, we’ll get the same answer either way.)

Let’s use [latex]MR = MC[/latex] this time.
\[ \begin{align*}
MR=& 300-0.04q\\
MC=& 30\\
300-0.04q=& 30\\
270=& 0.04q\\
q=& 6750
\end{align*} \]

The only critical point is at [latex]q = 6750[/latex]. Now we need to be sure this is a local max and not a local min. In this case, we’ll look to the graph of [latex]P(q)[/latex] – it’s a downward opening parabola, so this must be a local max. And since it’s the only critical point, it must also be the global max.

Profit is maximized when they sell 6750 ribbon winders.

Average Cost has critical points when Average Cost and Marginal Cost are equal.