Make a sketch of a rectangle with a line parallel to the width dividing the rectangle into two equal parts.
Let the length of the smaller rectangle be x and its width equal to y.
(the whole rectangle would be 2x by y)
given: 2xy = 225
y = 225/(2x)
length of fence = L
L = 4x + 3y
L = 4x + 675/2x = 4x + 337.5/x
dL/dx = 4 - 337.5/x^2
= 0 for a minimum of L
4 = 337.5/x^2
x^2 = 84.375
x = appr 9.19 ft
y = 18.37 ft
L = 91.86 ft as the minimum
A rectangular fenced enclosure of area 225 square feet is divided half into 2 smaller rectangles.
What is the minimum total material needed to build such an enclosure?
4 answers
three fences of length x
two fences of length y
A = xy = 225 so x = 225/y
p = total length = 2y + 3 x
we want to minimize p
p = 2 y +3(225/y)
dp/dy = 0 at max or min
dp/dy = 0 = 2 -675/y^2
2 y^2 = 675
y^2 = 337
y = 18.4
x = 225/18.4
so what is 2y+3x?
two fences of length y
A = xy = 225 so x = 225/y
p = total length = 2y + 3 x
we want to minimize p
p = 2 y +3(225/y)
dp/dy = 0 at max or min
dp/dy = 0 = 2 -675/y^2
2 y^2 = 675
y^2 = 337
y = 18.4
x = 225/18.4
so what is 2y+3x?
As always, in similar problems, the fence is evenly divided into lengths and widths.
So, you know right off that 2y=3x.
y(2y/3) = 225
y^2 = 675/2
y = 18.37
x = 12.23
So, you know right off that 2y=3x.
y(2y/3) = 225
y^2 = 675/2
y = 18.37
x = 12.23
Right!
Which apparently that isn't the right answer! I did pretty much exactly what @Damon and @Steve but my teacher marked it as wrong. What gives?????
Which apparently that isn't the right answer! I did pretty much exactly what @Damon and @Steve but my teacher marked it as wrong. What gives?????
3 answers