A rectangular fence has to be built from both wood and metal so that opposite parallel sides of the fence are made from the same type of material. The wood costs $10 per foot and the metal costs $20 per foot of the fence used. If you only have $400 to spend on material, what dimensions of the fence will enclose the largest area?
if the dimensions for metal and wood are m and w, then we know that
cost = 2(10w+20m)=400
area = m*w
= m(200-20m)/10
= 20m-2m^2
This is just a parabola with its vertex at m=5
So, the maximum area is with dimensions
metal=5 ft
wood=10 ft
To determine the dimensions of the fence that will enclose the largest area, we can use the concept of calculus to find the maximum area. Let's denote the length of the fence as L and the width as W.
To maximize the area of the fence, we'll first express the area as a function of one variable. In this case, we'll choose to express the area in terms of L. The width W can then be expressed as (400 - 10L)/30, as opposite parallel sides are made from the same type of material.
The area A of the rectangle is given by the formula A = L * W = L * (400 - 10L)/30.
To find the maximum area, we'll take the derivative of the area function with respect to L and set it equal to zero. Let's calculate the derivative of A:
dA/dL = (400 - 10L)/30 + (-10/30)L.
Setting dA/dL equal to zero:
(400 - 10L)/30 + (-10/30)L = 0.
Simplifying this equation:
(400 - 10L - 10L)/30 = 0,
(400 - 20L)/30 = 0,
400 - 20L = 0,
20L = 400,
L = 20.
Now, to find W, substitute the value of L into the expression for W:
W = (400 - 10(20))/30,
W = (400 - 200)/30,
W = 200/30,
W ≈ 6.67.
Therefore, the dimensions of the fence that will enclose the largest area are approximately L = 20 feet and W ≈ 6.67 feet.