Question

Sienna has 80 yards of fencing to enclose a rectangular area. Find the dimensions that maximize the enclosed area. What is the maximum area?

Answer 1

$2x+2y=80\ \ \ /:2\\\\2x:2+2y:2=80:2\\\\x+y=40\ \ \ /-x\\\\y=40-x\ \ \ (D_x:x\in(0;\ 40))$

$Area=xy\\\\substitute\ y=40-x\\\\Area=x(40-x)=-x^2+40\\(it's\ quadratic\ function\ where\ a=-1;b=40;c=0)\\\\vertex\ of\ parabola:p=\frac{-b}{2a}\to p=\frac{-40}{2\cdot(-1)}=\frac{-40}{-2}=20-it's\ max\\\\x=20\ then\ y=40-20=20\\\\Answer:dimensions\ of\ rectangular\ is\ 20\ yd\ \times\ 20\ yd,\\and\ area\ is\ 20^2=400\ yd^2.$