(1 point) A company has found that the relationship between the price and the demand for a particular product is given approximately by
The company also knows that the cost of producing the product is given by .
Find , the profit function.

Now use the profit function to do the following:
(A) Find the average of the values of all local maxima of .
Note: If there are no local maxima, enter -1000 .
Average of values
(B) Find the average of the values of all local minima of .
Note: If there are no local minima, enter -1000 .
Average of values
© Use interval notation to indicate where is concave up.
Note: Enter ‘I’ for , ‘-l’ for , and ’ ’ for the union symbol.
If you have extra boxes, fill each in with an ’ '.
Concave up:
( .

Ask by Rodriguez Bryant. in the United States

Mar 25,2025

Question

(1 point) A company has found that the relationship between the price and the demand for a particular product is given approximately by
The company also knows that the cost of producing the product is given by .
Find , the profit function.

Now use the profit function to do the following:
(A) Find the average of the values of all local maxima of .
Note: If there are no local maxima, enter -1000 .
Average of values
(B) Find the average of the values of all local minima of .
Note: If there are no local minima, enter -1000 .
Average of values
© Use interval notation to indicate where is concave up.
Note: Enter ‘I’ for , ‘-l’ for , and ’ ’ for the union symbol.
If you have extra boxes, fill each in with an ’ '.
Concave up:
( .

Ask by Rodriguez Bryant. in the United States

Mar 25,2025

UpStudy AI · Accepted Answer

We are given the profit function $$ P(x)=-0.13x^3+999x-870. $$ Step 1. Find the first derivative $$P'(x)$$: $$ P'(x)=-0.39x^2+999. $$ Critical points occur when $$P'(x)=0$$: $$ -0.39x^2+999=0 \quad \Longrightarrow \quad x^2=\frac{999}{0.39}. $$ Taking square roots gives $$ x=\pm\sqrt{\frac{999}{0.39}}. $$ Thus, there are two critical values: - $$x=\sqrt{\frac{999}{0.39}}$$ (approximately $$50.61$$) - $$x=-\sqrt{\frac{999}{0.39}}$$ (approximately $$-50.61$$). Step 2. Determine the nature of each critical point using the second derivative. Compute $$P''(x)$$: $$ P''(x) = \frac{d}{dx}(-0.39x^2+999) = -0.78x. $$ Evaluate $$P''(x)$$ at each critical point: - At $$x=\sqrt{\frac{999}{0.39}}$$: $$ P''\Bigl(\sqrt{\frac{999}{0.39}}\Bigr) = -0.78\sqrt{\frac{999}{0.39}} < 0, $$ so this point is a local maximum. - At $$x=-\sqrt{\frac{999}{0.39}}$$: $$ P''\Bigl(-\sqrt{\frac{999}{0.39}}\Bigr) = -0.78\Bigl(-\sqrt{\frac{999}{0.39}}\Bigr) = 0.78\sqrt{\frac{999}{0.39}} > 0, $$ so this point is a local minimum. Step 3. Answer the parts: (A) The local maximum occurs at $$x=\sqrt{\frac{999}{0.39}}$$ (approximately $$50.61$$). Since this is the only local maximum, the average of the $$x$$-values of all local maxima is $$ \boxed{50.61}. $$ (B) The local minimum occurs at $$x=-\sqrt{\frac{999}{0.39}}$$ (approximately $$-50.61$$). Since this is the only local minimum, the average of the $$x$$-values of all local minima is $$ \boxed{-50.61}. $$ (C) The concavity is determined by the sign of $$P''(x)=-0.78x$$: $$ P''(x)>0 \quad \Longrightarrow \quad -0.78x>0 \quad \Longrightarrow \quad x<0. $$ Thus, $$P(x)$$ is concave up on the interval $$ \boxed{(-l,0)}. $$ All answers are provided in the boxes as required. (A) Average of $$ x $$ values of local maxima: 50.61 (B) Average of $$ x $$ values of local minima: -50.61 (C) Concave up: $$ (-\infty, 0) $$

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