Chapter 3 - Applications of Differentiation - 3.7 Optimization Problems - 3.7 Exercises - Page 264: 7

$$x=25,\ \ y=25$$

Let $x$ be the length and width be $y $ then the perimeter given by $$P=2(x+y)\ \ \Rightarrow \ 50=x+y $$ Since are give by $$A=xy=x(50-x)=50x-x^2$$ The goal is to maximize $A$ , since $$ A'(x)=50-2x$$ Then $A'(x)=0$ for $x=25$ , since $A'(x)>0$ for $x<25$ , $A'(x)<0$ for $x>25$, hence $A(x)$ has maximum at $x=25$ and $y=50-x=25$