Chapter 12 - Section 12.2 - Applications of Maxima and Minima - Exercises - Page 886: 1

$25$

We are given that $P=xy~~~(a)$ and $x+y=10~~~(b)$ or, $y=10-x~~~(c)$ So, $P=x(10-x)=10x-x^2$ For maximise, we must have $\dfrac{dP}{dx}=0$ or, $10-2x=0 \\ 2x=10 \\ x=5$ Now, equation $c$ becomes: $y=10-5=5$ Thus, our equation (a) becomes: $P_{max}=(5)(5)=25$