Chapter 12 - Section 12.2 - Applications of Maxima and Minima - Exercises - Page 886: 9

$25$

Here, we have: $2(x+y)=20~~~~(a)$ where, $x$ and $y$ represents length and breadth. Next, $x+y=10 \implies x=10-y~~~(b)$ So, the area can be written as: $A=xy=(10-y)y=10y-y^2$ For maximum, we must have $\dfrac{dA}{dy}=0$ or, $10-2y=0 \\ 2y=10 \\ y=5$ Now, equation $b$ becomes: $x=10-5=5$ Thus, the maximum area is: $F_{max}=(5)(5)=25$