Some cowboys plan to build a rectangular corral in a canyon. They have 500 feet of fencing to use for three sides of the rectangle (the canyon wall is used for the 4th side).
a) What is the maximum area that they can fence in?
b) What are the dimensions of the rectangle?
let the side parallel to the canyon wall be y ft
let each of the other two shorter and equal sides be x
2x + y = 500
y = 500-2x
area = xy
= x(500-2x)
= 500x - 2x^2
since you labeled it Calculus
d(area)/dx = 500 - 4x
= 0 for a max of area
4x = 500
x = 125
when x = 125 , y = 500 - 2(125) = 250
max area = xy = 125(250) = 31250 ft^2 when the longer side is 250 ft and the two shorter sides are 125 ft each
Some cowboys plan to build a rectangular corral in a canyon. They have 200 feet of fencing to use for three sides of the rectangle (the canyon wall is used for the 4th side).
a) What is the maximum area that they can fence in?
b) What are the dimensions of the rectangle?
My bad.
To find the maximum area that can be fenced in, we need to determine the dimensions of the rectangle.
Let's assume the length of the rectangle is L and the width is W.
a) To maximize the area, we need to maximize the product of the length and width: A = L x W.
b) We have three sides of fencing, so the perimeter is: P = 2L + W.
Given that we have 500 feet of fencing, we can write the equation: 2L + W = 500.
To find the maximum area, we need to eliminate one variable from the equation above. We can express L in terms of W by rearranging the equation: L = (500 - W) / 2.
Substituting this value of L into the area equation, we get: A = [(500 - W) / 2] x W.
Now, we have the area equation in terms of just one variable, W. To find the maximum area, we need to find the value of W that maximizes the area.
We can do this by either graphing the equation (A vs. W) and finding the maximum point or by using calculus to find the critical points.
Using calculus, we take the derivative of A with respect to W, set it equal to zero, and solve for W.
dA/dW = (500 - W) / 2 - W / 2 = 0.
Simplifying this equation, we get: 500 - W - W = 0, which gives us: 500 - 2W = 0.
Solving for W, we find: W = 250.
Plugging this value of W back into the perimeter equation, we can find the corresponding value of L: 2L + 250 = 500, which leads to L = 125.
So, the dimensions of the rectangle that maximize the area are 125 feet in length and 250 feet in width.
To find the maximum area, we plug these values into the area equation: A = 125 x 250 = 31,250 square feet.
Therefore, the maximum area that can be fenced in is 31,250 square feet.