2.9: Optimization

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.

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.

Answer: dimensions of enclosure?

Objective: minimize cost

Objective function: [latex]C=14x+7y+7x+7y=21x+14y[/latex]

We must rewrite the objective function as a function of one variable, i.e., eliminate one of the two variables.

Since we have a condition on the area size, we have [latex]A = xy = 600[/latex] where [latex]A[/latex] is the area. Solving for [latex]x[/latex] we get [latex]x=\frac{600}{y}[/latex]. We then substitute the value of [latex]x[/latex] into [latex]C[/latex]:

\[C=21\left(\frac{600}{y}\right)+14y=\frac{12600}{y}+14y=12600y^{-1}+14y\]

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

First, the domain of [latex]C[/latex]: [latex]y[/latex] represents the length of the enclosure, so [latex]y>0[/latex].

Then we need to find the critical numbers, which will give us the potential local minima or maxima for the cost. So we need to find out where [latex]C'(y)[/latex] is zero or undefined. Since

\[C'(y)=12600(-1)y^{-1-1}+14=-12600y^{-2}+14=-\frac{12600}{y^2}+14\]

we have that [latex]C'(y)[/latex] is undefined for [latex]y = 0[/latex], which is not in the domain and so [latex]y=0[/latex] is not a critical number. We also have that [latex]C' (y)= 0[/latex] when

\[
\begin{align*}
-\frac{12600}{y^2}+14&=0\\\\
\frac{-12600+14y^2}{y^2}&=0\\\\
-12600+14y^2&=0\\\\
y^2&=\frac{12600}{14}=900\\\\
y&=\pm 30
\end{align*}
\]

Since only [latex]y=30[/latex] is in the domain, it is the only critical number.  Therefore [latex]y=30[/latex] gives the only potential maximum or minimum cost.

We now use the second derivative test to determine the answer, if possible:

\[
C”(y)=-12600(-2)y^{-2-1}+0=25200y^{-3}=\frac{25200}{y^3} \Rightarrow C”(30)\gt 0
\]

so [latex]y=30[/latex] gives a local minimum.

Since this is the only critical point in the domain, this must give the global minimum for the cost.

Going back to our constraint function, we can find that when [latex]y = 30[/latex], [latex]x = \frac{600}{30}=20[/latex].

Therefore, the dimensions of the enclosure that minimize the cost are 20 feet by 30 feet, where the more expensive side is 20 feet long.

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 and what will be the maximum revenue?

Answer:  price [latex]\$p[/latex] per ticket?

Objective: maximize revenue [latex]R=p\cdot x[/latex], where [latex]p[/latex] is the price per ticket, and [latex]x[/latex] is the number of tickets sold.

To find where the revenue is at the maximum, we have to rewrite it as a function in one variable, so we must eliminate either [latex]p[/latex] or [latex]x[/latex]. Since we are looking for the price, let’s eliminate [latex]x[/latex]. This means that we have to rewrite [latex]x[/latex] as a function of [latex]p[/latex].

We know that the price and the number of tickets are linearly related because the number of tickets sold changes at the constant rate in relation to change in price. So [latex]x(p)=mp+b[/latex] where

\[
m=\frac{\text{change in output}}{\text{change in input}}=\frac{x_2-x_1}{p_2-p_1}=\frac{1200-1150}{50-55}=-10
\]

Therefore, [latex]x(p)=-10p+b[/latex]. To find the value of [latex]b[/latex], we can substitute one of the price points:

\[
x(50)=1200\Rightarrow -10(50)+b=1200\Rightarrow b=1700
\]

Therefore, [latex]x=-10p+1700[/latex] and so [latex]R(p)=p(-10p+1700)=-10p^2+1700p[/latex].

We now have to determine the domain of [latex]R(p)[/latex]. Since [latex]p[/latex] represents the price, [latex]p\ge 0[/latex]. Moreover, since this problem only discussed raises in price from the starting price of $50 per ticket, we must have [latex]p\ge 50[/latex]. In addition, the quantity sold must be non-negative, and so [latex]x(p)\ge 0[/latex]. This means that [latex]-10p+1700\ge 0[/latex] and so [latex]p\le 170[/latex]. Therefore, the domain for the revenue function is [latex]50\le p\le 170[/latex].

Now, we can find the maximum of the revenue function by finding critical numbers.

[latex]R'(p)=1700-20p[/latex], so [latex]R'=0[/latex] when [latex]p = 85[/latex], which is in the domain of the revenue, and the derivative is never undefined. Therefore, [latex]p = 85[/latex] is a critical number and gives potential local minimum or maximum for the revenue.

Using the second derivative test, [latex]R''(p)=-20[/latex], which is negative for any value of [latex]p[/latex], so the critical number [latex]p = 85[/latex] gives a local maximum for the revenue. Since it is the only critical number, we can also conclude that it gives the global maximum.

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

Suppose we want to maximize profit.

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

But remember that Profit = Revenue – Cost or [latex]P=R-C[/latex]. So [latex]P'=R'-C'[/latex]. That is, the derivative of the profit function is marginal revenue minus marginal cost.

Now let’s find the critical points for the profit – those will be where [latex]P'=0[/latex] or is undefined. We have [latex]P'=0[/latex]  when [latex]R'-C'= 0[/latex], or where [latex]R'= C'[/latex].

Therefore, the profit has critical points when the marginal revenue and marginal cost are equal.

 

A company sells [latex]x[/latex] ribbon winders per year at [latex]\$p[/latex] per ribbon winder. The demand function for ribbon winders is given by: [latex]p=200-0.02x[/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.

Answer: quantity [latex]x=?[/latex]

Objective: maximize profit

Objective function: [latex]P=R-C[/latex] where [latex]R[/latex] is the revenue from selling and [latex]C[/latex] is the cost of producing [latex]x[/latex] ribbon winders.

To determine the revenue in terms of the quantity sold, we can use

\[
R(x)=x\cdot p(x)=x(200-0.02x)
\]

To determine the cost, we can use

\[
C(x)=30x+9000
\]

Therefore,

\[
\begin{align*}
P(x)&=R(x)-C(x)\\
&=x(200-0.02x)-(30x+9000)\\
&=200x-0.02x^2-30x-9000\\
&=-0.02x^2+170x-9000
\end{align*}
\]

Now we have two choices. We can find the critical points of profit by taking the derivative of [latex]P(x)[/latex] directly, or we can find [latex]R'[/latex] and [latex]C'[/latex] and set them equal. (Naturally, we’ll get the same answer either way.)
Let’s use [latex]R' = C'[/latex] this time.

\[
\begin{align*}
R’ & = 200-0.04x\\
C’ & = 30\\
200-0.04x & = 30\\
170 & = 0.04x\\
x & =4250
\end{align*} \]

The only critical point is at [latex]x =4250[/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(x)[/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.

Therefore, profit is maximized when they sell 4250 ribbon winders.