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A retail grocer has a daily demand Y for a certain food
Chapter 4, Problem 179SE(choose chapter or problem)
A retail grocer has a daily demand 𝑌 for a certain food sold by the pound, where 𝑌 (measured
in hundreds of pounds) has a probability density function given by
\(f(y)=\left\{\begin{array}{ll}
3 y^{2}, & 0 \leq y \leq 1, \\
0, & \text { elsewhere. }
\end{array}\right.
\)
(She cannot stock over 100 pounds.) The grocer wants to order 100𝑘 pounds of food. She buys
the food at 6¢ per pound and sells it at 10¢ per pound. What value of 𝑘 will maximize her expected daily profit?
Equation Transcription:
Text Transcription:
f(y)=
3y^2, 0</=y</=1,
0, elsewhere.
Questions & Answers
QUESTION:
A retail grocer has a daily demand 𝑌 for a certain food sold by the pound, where 𝑌 (measured
in hundreds of pounds) has a probability density function given by
\(f(y)=\left\{\begin{array}{ll}
3 y^{2}, & 0 \leq y \leq 1, \\
0, & \text { elsewhere. }
\end{array}\right.
\)
(She cannot stock over 100 pounds.) The grocer wants to order 100𝑘 pounds of food. She buys
the food at 6¢ per pound and sells it at 10¢ per pound. What value of 𝑘 will maximize her expected daily profit?
Equation Transcription:
Text Transcription:
f(y)=
3y^2, 0</=y</=1,
0, elsewhere.
ANSWER:
Solution :
Step 1 of 1:
Let Y has a probability density function is given by
Our goal is:
We need find the value of k will maximize her expected daily profit.
Now we need find the value of k will maximize her expected daily profit.
Given she buys the food at 6¢ per pound and sells 10¢ per pound.
If Xk ; p=(10-b)x-b(k-x)