Chapter 12 - Section 12.2 - Applications of Maxima and Minima - Exercises - Page 886: 2

$200$

We are given that $P=xy~~~(a)$ and $x+2y=40~~~(b)$ or, $x=40-2y~~~(c)$ So, $P=(40-2y)(y)=40y-2y^2$ For maximise, we must have $\dfrac{dP}{dx}=0$ or, $40-4y=0 \\ 4y=40 \\ y=10$ Now, equation $c$ becomes: $x=40-(2)(10)=20$ Thus, our equation (a) becomes: $P_{max}=(20)(10)=200$