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Please help me with this problem.

Please help me with this problem.-example-1
User Annick
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We are given the revenue function $R(x)=x \cdot p(x)$, where $p(x)=300-0.25 x$ is the unit price. To find the maximum revenue, we need to find the value of $x$ that maximizes $R(x)$.

\begin{align*}
R(x) &= x \cdot p(x) \\
&= x \cdot (300 - 0.25x) \\
&= 300x - 0.25x^2
\end{align*}

To find the maximum revenue, we need to find the value of $x$ that maximizes $R(x)$. We can do this by taking the derivative of $R(x)$ with respect to $x$, setting it equal to zero, and solving for $x$.

\begin{align*}
R'(x) &= 300 - 0.5x \\
0 &= 300 - 0.5x \\
0.5x &= 300 \\
x &= 600
\end{align*}

Therefore, the maximum revenue is achieved when $x=600$. To find the maximum revenue, we can substitute this value into the revenue function:

\begin{align*}
R(600) &= 600 \cdot (300 - 0.25 \cdot 600) \\
&= 600 \cdot 150 \\
&= \boxed{90,000}
\end{align*}

Therefore, the maximum revenue for the bicycle shop is $90,000.



Let me know if this is correct:)
User Ganesh Kudva
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