Question

Can you check to see if my work is correct?

Answer 1

The Solution.

`P=-(1)/(250)T^2+2.8T-394`

To obtain the maximum value of T, we differentiate with respect to x and equate to zero.

`\begin{gathered} (dP)/(dT)=2(-(1)/(250)T)+2.8\text{ =0} \\ \\ -(1)/(125)T+2.8=0 \\ \\ -(T)/(125)=-2.8 \\ \text{Cross multiplying, we get} \\ T=125*2.8=350^oF \end{gathered}`

To get the maximum value of P, we shall substitute 350 for T in the given function.

`\begin{gathered} P=-(1)/(250)(350^2)+2.8(350)-394 \\ \\ P=-490+980-394 \\ P=96\text{ percent} \end{gathered}`

The correct answer is T = 350 degrees Fahrenheit , and P = 96 percent.