# Rope and wire are arranged in the shape of a rectangle. Two parallel sides of the rectangle are made from the rope which costs \$2.00 per foot. The other two sides are made from the wire which costs \$3.00 per foot. Find the dimensions of the rectangle enclosing the most area which costs \$100.00.

Looking for: max area
area= w x r
100=3w + 2r
w=100/3 - 2/3r
a(r)= (100/3 - 2/3r)r
a`(r)=100/3 - 4/3r
r=25
100=3w + 2r or 100=3w + 2(25)
w=50/3

## To find the dimensions of the rectangle enclosing the most area, we need to maximize the area function, which is given by A = w * r, where w represents the width and r represents the length of the rectangle.

We are given that the cost of the rope is \$2.00 per foot, and the cost of the wire is \$3.00 per foot. So, the cost function is given by C = 3w + 2r.

We are also given that the cost should be \$100.00. So, we can set up the equation C = 100 as follows:
100 = 3w + 2r

From this equation, we can solve for w in terms of r:
w = (100/3) - (2/3)r

Now, we can substitute this value of w into the area function to get a function for the area in terms of r:
A(r) = [(100/3) - (2/3)r] * r

To find the maximum area, we can find the critical points of A(r) by taking the derivative and setting it equal to zero:
A'(r) = (100/3) - (4/3)r

Setting A'(r) equal to zero, we get:
(100/3) - (4/3)r = 0
100 - 4r = 0
4r = 100
r = 25

Now that we have the value of r, we can substitute it into the equation for w:
w = (100/3) - (2/3)(25)
w = 50/3

Therefore, the dimensions of the rectangle enclosing the most area, with a cost of \$100.00, are:
Width (w) = 50/3 feet
Length (r) = 25 feet