SOLUTION: A farmer wants to make a retangular 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 for
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Question 196948: A farmer wants to make a retangular 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" Answer by ankor@dixie-net.com(22740) (Show Source):
You can put this solution on YOUR website! 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 =
x =
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.
