Question

Suppose the Sunglasses Hut Company has a profit function given by P(q) = -0.01q^2 + 3q– 37, whereq is the number of thousands of pairs of sunglasses sold and produced, and P(q) is the total profit, inthousands of dollars, from selling and producing a pairs of sunglasses.A) How many pairs of sunglasses (in thousands) should be sold to maximize profits? (If necessary, round youranswer to three decimal places.)Answer:thousand pairs of sunglasses need to be sold.B) What are the actual maximum profits (in thousands) that can be expected? (If necessary, round youranswer to three decimal places.)Answer:thousand dollars of maximum profits can be expected.

Answer 1

For item a) we need to find the maxima of a function. If we take the derivative of a function, when its equal to 0, it could be a maxima or a minima

Let's take the derivative:

$P(q)=-0.01q^2+3q-37$

By using the derivative rules:

$\begin{gathered} P^{\prime}(q)=-0.01\cdot2\cdot q+3 \\ P^{\prime}(q)=-0.02q+3 \end{gathered}$

Now, by making the derivative equal 0, we can find the critic points:

$\begin{gathered} \text{If}P^{\prime}(q)=0 \\ \text{Then:} \\ 0=-0.02q+3 \end{gathered}$

And solve for q:

$\begin{gathered} \frac{-3}{-0.02}=q \\ q=150 \end{gathered}$

We can infer that the function P(q) has a critical point in q = 150. To know what is the nature of that critical point, we can take the second derivative:

$P^"(q)=-0.02$

Since the second derivative is negative, the point q = 150 is a maximum.

Thus, the answer to question a is

Answer: 150 thousand pairs of sunglasses needs to be sold.

Now to part b)

Since the maximum profit will occur when q = 150, to know the maximum profit, we need to evaluate P(q) when q = 150:

$P(150)=-0.01(150)^2+3\cdot150-37$

Solving:

$\begin{gathered} P(150)=-225+450-37 \\ P(150)=188 \end{gathered}$

The answer to b is:

Answer: 188 thousand dollars of maximum profit can be expected