Questions: QUESTION (2 pts) A firm that manufactures special bicycles has the following profit function (in ): π(f, w)=3 f^2-14 f w+5 w^2+350 f Where f denotes the number of frames and w denotes the number of wheels. The firm doesn't want spare frames or wheels left over at the end of the production run (i.e. the firm wants to use all the production of frames and wheels). How many frames f and wheels w maximize the profit (in ) when satisfying the firm condition? (Round your answers to the nearest unit if needed) f= w=

QUESTION (2 pts) A firm that manufactures special bicycles has the following profit function (in ): π(f, w)=3 f^2-14 f w+5 w^2+350 f Where f denotes the number of frames and w denotes the number of wheels. The firm doesn't want spare frames or wheels left over at the end of the production run (i.e. the firm wants to use all the production of frames and wheels). How many frames f and wheels w maximize the profit (in ) when satisfying the firm condition? (Round your answers to the nearest unit if needed) f= w=
Transcript text: QUESTION (2 pts) A firm that manufactures special bicycles has the following profit function (in \$): \[ \pi(f, w)=3 f^{2}-14 f w+5 w^{2}+350 f \] Where $f$ denotes the number of frames and $w$ denotes the number of wheels. The firm doesn't want spare frames or wheels left over at the end of the production run (i.e. the firm wants to use all the production of frames and wheels). How many frames $f$ and wheels $w$ maximize the profit (in \$) when satisfying the firm condition? (Round your answers to the nearest unit if needed) \[ \begin{array}{l} f= \\ w= \end{array} \] $\square$
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Solution

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Solution Steps

Step 1: Define the Profit Function

The profit function is given by

\[ \pi(f, w) = 3f^2 - 14fw + 5w^2 + 350f \]

Step 2: Substitute the Constraint

Assuming the firm wants to use all frames and wheels, we set \(w = f\). Thus, the profit function simplifies to

\[ \pi(f) = 3f^2 - 14f^2 + 5f^2 + 350f = -6f^2 + 350f \]

Step 3: Find the Critical Points

To find the maximum profit, we take the derivative of the simplified profit function with respect to \(f\):

\[ \frac{d\pi}{df} = 350 - 12f \]

Setting the derivative equal to zero to find critical points:

\[ 350 - 12f = 0 \implies f = \frac{350}{12} = \frac{175}{6} \]

Step 4: Determine the Corresponding Value of \(w\)

Since we assumed \(w = f\), we have

\[ w = \frac{175}{6} \]

Final Answer

The values of frames and wheels that maximize the profit are

\[ \boxed{f = \frac{175}{6}, \quad w = \frac{175}{6}} \]

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