Answer:
The answer is below
Step-by-step explanation:
Let x represent the length of the rectangular pen and y represent the width of the rectangular pen.
One side of the pen uses the barn wall, hence the remaining side is 64 m, hence:
x + 2y = 64
x = 64 - 2y (1)
The area of the wall is the product of the length and the width, hence:
Area (A) = length * width = xy = (64 - 2y)y = 64y - 2y²
A = 64y - 2y²
The maximum area of the dog house is gotten when the derivative of the area is equal to 0. Hence:
A' = 64 - 4y = 0
64 - 4y = 0
4y = 64
y = 16 m
Put y = 16, in equation 1 to find x:
x = 64 - 2(16) = 32 m
x = 32 m
Therefore the pen has a length of 32 m and width of 16 m.
Answer:
A(max) = 512 m²
Dimensions:
x = 32 m
y = 16 m
Step-by-step explanation:
Area of the rectangular pen:
A(p) = x*y x and y sides of the rectangle. A barn will be used in place of a fence; let´s say that only one x side will be fenced, then the perimeter of the rectangle is:
p = x + 2*y 64 = x * 2*y y = ( 64 - x ) / 2
A(p) = x*y
Area as a function of x
A(x) = x * ( 64 - x )/ 2
A(x) = ( 64*x - x² ) /2
Tacking derivatives on both sides of the equation
A´(x) = ( 64 -2*x )/2
A´(x) = 0 64 - 2*x = 0 2*x = 64 x = 32 m
y = ( 64 - x ) / 2 y = ( 64 - 32 )/2 y = 16 m
A(max) = 32 * 16 = 512 m²
To check if x = 32 will bring a maximum to function A(x)
We take the second derivative
A´´(x) = - x/4 A´´(x) < 0 then x = 32 is a local maximum
