Question 196948
A farmer wants to make a rectangular corral along the side of a large barn
 and has enough materials for 60m of fencing. Only three sides must be fenced,
 since the barn wall will form the fourth side. 
What width of rectangle should the farmer use so that the maximum area is enclosed? 
Use "let statements"
:
Let x = width of the rectangle
let L = length of the rectangle
:
Because there are only 3 sides required, the perimeter will be:
L + 2x = 60
therefore;
L = (60-2x)
:
Area = x * L
replace L with (60-2x)
A = x(60-2x)
A = 60x - 2x^2
A quadratic equation:
y = -2x^2 + 60x
:
Find the max area width by finding the axis of symmetry; x = -b/(2a)
In this equation a=-2; b=60
x = {{{(-60)/(2*-2)}}}
x = {{{(-60)/(-4)}}}
x = +15 meters is the width for max area
:
;
You can use x=15 in the area equation to find the max area, if you want to.