Question 248944
Word Problem: A rancher can afford 300 feet of fencing to build a corral that's divided into two equal rectangles.
 What dimensions will maximize the corral's area?
I'm given a picture where the length is 2x and the width is y
:
I have no idea why you would use distance formula, perhaps he meant the perimeter formula
:
The fence equation
2(2x) + 3y = 300
4x + 3y = 300
3y = (300 - 4x)
y = 100 - {{{4/3}}}x
:
Area = L * W
A = 2x * y
Replace y with(100-{{{4/3}}}x)
A = 2x(100-{{{4/3}}}x)
A = 200x - 2x({{{4/3}}}x)
A = 200x -{{{8/3}}}x^2
:
A quadratic equation, the max area occurs at the axis of symmetry
The formula for that: x = -b/(2a)
In this equation a=-8/3; b=200
x = {{{-200/(2(-8/3))}}}
x = {{{-200/(-16/3)}}}
x = -200 * {{{-3/16}}}
x = +37.5
:
Find y:
y = 100 - {{{4/3}}}x
y = 100 - {{{4/3}}}*37.5
y = 100 - 50
y = 50 ft
;
Since the dimensions are are 2x by y, max area occurs when:
2(37.5) = 75 ft is the length and 50 ft is the width
:
Check the perimeter (fence length)
2(75) + 3(50) =
150 + 150 = 300
;
;
C