a farmer wants to make a rectangular enclosure using a wall as one side and 120 m of fencing for the other three sides.
a) express the area in terms of x and state the domain of the area function
b)find the value of x that gives the greatest area.
my teacher didn't explain how to do this so im not sure, please help!
Oooppps. If you are in precal, you may not know derivative yet. In that case, graph A vs s, look for the max A, read where s is, then solve for l knowing s.
perimeter= 2s + l
120=2s + l
area= sl= s(120-2s)
Take the deriviative of area with respect to s, set to zero, and find s,l.
Sure, I'd be happy to help!
To solve this problem, let's start by visualizing the rectangular enclosure.
The farmer wants to use a wall as one side, so we can think of the rectangular enclosure as having one side of length x (which is the length of the wall), and the other two sides are each equal to 120 m (the remaining 3 sides of fencing).
a) Expressing the area in terms of x:
The formula to find the area of a rectangle is A = length * width. In this case, the length is x, and the width is 120 m.
Therefore, the area of the rectangular enclosure can be expressed as A = x * 120.
The domain of the area function:
To determine the domain of the area function, we need to consider the constraints of the problem.
Given that the farmer has 120 m of fencing for the other three sides, the sum of the three sides must be equal to 120 m. Since two of the sides have a length of 120 m each, the remaining side (which is the length of the wall) cannot exceed 120 m.
So, the domain of the area function would be x ≤ 120, since the length of the wall cannot exceed 120 m.
b) Finding the value of x that gives the greatest area:
To find the value of x that gives the greatest area, we need to maximize the area function A = x * 120 within the given domain.
We can achieve this by finding the critical points of the function, which occur when the derivative of the function is equal to zero.
Let's differentiate the area function with respect to x: dA/dx = 120.
Setting the derivative to zero and solving for x, we get:
120 = 0
x = 0
However, this value is not valid in the context of the problem since the length of the wall cannot be zero.
So, in this case, we need to check the endpoints of the domain, which are x = 0 and x = 120.
Substituting x = 0:
A = 0 * 120 = 0
Substituting x = 120:
A = 120 * 120 = 14400
Therefore, the value of x that gives the greatest area is x = 120, which corresponds to a square enclosure.