7) A rectangular pen can be constructed using the side of a barn as one boundary and 5500 ft. of fence to make the other three sides. Find the length and width of the rectangle to make the largest area.
Hint: Draw a picture- and label the sides of the rectangle. And solve the problem.
let each of the two equal sides be x
let the single side by y
then y+2x = 5500
y = 5500 - 2x
area = xy
= x(5500-2x)
= -2x^2 + 5500
Since you labeled it "college" math, I will assume you know Calculus
d(area)/dx = -4x + 5500
= 0 for a max of area
4x = 5500
x = 1375
y = 5500-2(1375) = 2750
the pen is 1375 ft by 2750 ft, with the side of 2750 parallel to the barn.
To find the dimensions of the rectangle that will result in the largest area, we can use calculus. Let's start by drawing a rectangle as described in the problem:
```
__________________
| w |
| |
| barn |
| |
|________________|
```
Let's label the width of the rectangle as "w" and the length as "l". We know that one side of the pen is already defined by the barn, so the three sides that need to be fenced have a combined length of 5500 ft.
Given that the fenced sides have a length of 5500 ft, we can express the total perimeter of the rectangle as:
Perimeter = length + 2 * width
5500 = l + 2w
Now, we need to express the area of the rectangle in terms of l and w. The area of a rectangle is given by:
Area = length * width
A = l * w
We want to find the dimensions that maximize the area A.
To solve the problem, we will use the fact that when a function reaches a maximum (or minimum) value, its derivative is equal to zero. So, let's take the derivative of the area function A with respect to l:
dA/dl = w
Now, let's write the equation for the perimeter in terms of l:
l = 5500 - 2w
Substitute the value of l in the equation for the area:
A = (5500 - 2w) * w
Now, we can find the derivative of A with respect to w:
dA/dw = 5500 - 4w
To find the value of w that maximizes the area, set dA/dw equal to zero and solve for w:
5500 - 4w = 0
4w = 5500
w = 1375 ft
Now that we have the value of w, we can substitute it back into the equation for l:
l = 5500 - 2(1375)
l = 2750 ft
Therefore, the length and width of the rectangle that result in the largest area are 2750 ft and 1375 ft, respectively.