The Solution.
[tex]P=-\frac{1}{250}T^2+2.8T-394[/tex]
To obtain the maximum value of T, we differentiate with respect to x and equate to zero.
[tex]\begin{gathered} \frac{dP}{dT}=2(-\frac{1}{250}T)+2.8\text{ =0} \\ \\ -\frac{1}{125}T+2.8=0 \\ \\ -\frac{T}{125}=-2.8 \\ \text{Cross multiplying, we get} \\ T=125\times2.8=350^oF \end{gathered}[/tex]
To get the maximum value of P, we shall substitute 350 for T in the given function.
[tex]\begin{gathered} P=-\frac{1}{250}(350^2)+2.8(350)-394 \\ \\ P=-490+980-394 \\ P=96\text{ percent} \end{gathered}[/tex]
The correct answer is T = 350 degrees Fahrenheit , and P = 96 percent.
