A rectangular dog pen is to be constructed using a barn wall as one side and 60 meters of fencing for the other three sides. Find the dimensions of the pen that maximize the pen's area.
let the length be y m (the single side)
let the width be x m
so we know that 2x + y = 60
or y = 60 - 2x
Area = xy
= x(60-2x)
= -2x^2 + 60x
At this point I don't if you know Calculus or not.
If you do, then
d(Area)/dx = -4x + 60
= 0 for a max of area
-4x + 60 = 0
x = 15
then y = 30
if you don't take Calculus, we have to complete the square
area = -2(x^2 - 30x)
= -2(x^2 - 30x + 225 - 225)
= -2((x-15)^2 - 225)
= -2(x-15)^2 + 450
so the area is a maximum when x = 15
and then y = 60-30 = 30
by either method,
the width has to be 15 m, and the length has to be 30 m for a maximum area of 15(30) or 450 m^2
(LULU's answer only gives an area of 400 )
Why did the dog bring a tape measure to the farm? Because he was "dog-termined" to find the best dimensions for his pen!
Let's calculate the dimensions that maximize the area of the pen.
Let's assume the length of the barn wall is 'x' meters. This means the length of the other two sides would be (60 - x)/2 meters each since there are three equal sides.
The area of the pen can be calculated by multiplying the length and width. So, the area, A, can be expressed as:
A = x * (60 - x)/2
To find the maximum area, we can take the derivative of A with respect to x and set it equal to 0. Let's do some math to solve it.
First, we rewrite the equation:
A = (60x - x^2)/2
Taking the derivative:
dA/dx = (60 - 2x)/2
Setting it equal to 0:
(60 - 2x)/2 = 0
Simplifying:
60 - 2x = 0
-2x = -60
x = -60/-2
x = 30
Therefore, the length of the barn wall should be 30 meters. Plugging this into the equation for the width, we get:
(60 - x)/2 = (60 - 30)/2 = 30/2 = 15
So, the dimensions that maximize the area of the pen are 30 meters (length of the barn wall) and 15 meters (width).
Keep in mind, this solution assumes the barn wall is the longest side of the pen. If the barn wall is meant to be the width, you can exchange the length and width in the solution. Now that's a mathematical dog-tale!
To find the dimensions that maximize the pen's area, we need to maximize the area of the rectangle.
Let's denote the length of the rectangle as L and the width as W. Since the barn wall serves as one side, the perimeter of the pen (which is also the total length of the fence) is given by:
P = W + 2L
We also know that the total length of the fence should be 60 meters, so we can write the equation:
W + 2L = 60
To find the dimensions that maximize the area, we need to express the area of the rectangle in terms of a single variable, W or L. The area of a rectangle is given by:
A = L * W
We can substitute the value of W from the first equation into this expression to get:
A = L * (60 - 2L)
Now, we can simplify the equation:
A = 60L - 2L^2
To maximize the area, we can take the derivative of the equation with respect to L, set it equal to zero, and solve for L.
dA/dL = 60 - 4L = 0
60 = 4L
L = 15
Now that we have L = 15, we can substitute it back into the original equation to find W:
W + 2L = 60
W + 2 * 15 = 60
W + 30 = 60
W = 30
Therefore, the dimensions of the pen that maximize the area are length = 15 meters and width = 30 meters.