A veterinarian uses 300 feet of chain-link fencing to enclose a rectangular region and also to subdivide the region into two smaller rectangular regions by placing a fence parallel to one of the sides, as shown in the figure.
(a) Write the width w as a function of the length l.
w = _____
(b) Write the total area A as a function of l.
A = _____
(c) Find the dimensions that produce the greatest enclosed area.
_____ ft (smaller value)
_____ ft (larger value)
Sorry, but I did not post my question twice.
To solve this problem, we need to set up some equations and then maximize the area function by finding its critical points.
(a) Write the width w as a function of the length l:
Let the length of the rectangular region be l and the width be w. As given in the problem, we are dividing the rectangular region into two smaller regions by placing a fence parallel to one of the sides. This means that the height of the rectangular region will be divided into two equal parts, with each smaller region having a height of w/2.
Since we are using 300 feet of fencing, the total length of the two lengths (l) and the single width (w) is equal to 300 ft:
2l + w = 300
From this equation, we can solve for w:
w = 300 - 2l
(b) Write the total area A as a function of l:
The area of the large rectangular region is given by A = l * w. Substituting the value of w from the previous equation, we get:
A = l * (300 - 2l)
Simplifying this expression, we get:
A = 300l - 2l^2
(c) Find the dimensions that produce the greatest enclosed area:
To find the dimensions that produce the greatest enclosed area, we need to find the critical point of the area function A(l). We take the derivative of A(l) with respect to l and set it equal to zero:
dA/dl = 300 - 4l
Setting dA/dl = 0, we have:
300 - 4l = 0
4l = 300
l = 300/4
l = 75
Substituting this value back into the equation for w, we can find the value of w:
w = 300 - 2(75)
w = 300 - 150
w = 150
Therefore, the dimensions that produce the greatest enclosed area are:
75 ft (smaller value)
150 ft (larger value)