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Question

John has 300 feet of lumber to frame a rectangular patio (the perimeter of a rectangle is 2 times length plus 2 times width). He wants to maximize the area of his patio (area of a rectangle is length times width). What should the dimensions of the patio be, and show how the maximum area of the patio is calculated from the algebraic equation.

Show clearly the algebraic steps which prove your dimensions are the maximum area which can be obtained. Use the vertex formula to find the maximum area.

Answer:

Show work in this space.

Answer 1

2W + 2L = 300
L = 150 - W

A = LW = 150W - W^2
dA/dW = 150 - 2W

Max at dA/dW = 0

2W = 150
W=75, L=75
A(max) = 5625

Answer 2

Differential calculus will be used here.

Let L denote the length and B denote the breadth of the patio.

We are given that 300 feet of lumber is used to make the patio, which means that the perimeter is 300 feet.

2*(L+B) = 300
=> L + B = 150.........................Equation 1

Area of the patio = L * B.........................Equation 2

We need to maximize area of the lawn. We substitute for L from equation 1 in equation 2. We get:

Area = B * (150 - B) = 150B - B^2.........................Equation 3

For area to be maximum, we need to find the local maximum which we do by differentiating equation 3 with respect to B. We get:

dArea / dB = 150 - 2B

For a local maxima/minima, dArea / dB = 0

Hence, B = 75.

For a local maxima, the second differential of Area with respect to B will be negative at 75.

d^2Area / dB^2 = -2, which means that the point is a maxima.

Hence, the width of the patio should be 75 feet.

Substituting for B in equation 1, we get Length of the patio to be 75 feet. Hence, the optimum size would be a square with dimensions 75 feet by 75 feet.

Answer 3

Call the length L and the width W.

Write down a formula which relates the total length of wood to L and W. Now convert this to a formula which shows L as a function of W.

Write down a formula for area A based on L and W. Substitute the value for L you worked out earlier into this, to get a formula for A based only on W.

Differentiate with respect to W and set this equal to zero to find the value of W which gives maximum area.

Using this value of W, calculate L and A.

Answer 4

the area of the patio is xy =S

the perimeter of the patio 300 = 2x +2y or dividing by 2
150=x+y
y=150-x

and the area is x(150-x) = S =X^2-150x

you search the maximum of this

dS/dx = 2x-150 maximum then 2x-150 =0 x=75

The result is a quadrat x=y = 75 feet with area s =75^2 =5625m^2

Answer 5

x=width
y=length

Perimeter:
2x+2y = 300
y = 150-x

Area:
A=x*y
=x*(150-x)
=150x-x^2

dA/dx = 150 -2x = 0 =====> Max. area
2x = 150
x = 75 ft.

if x = 75ft.,

y = 150 - 75 = 75ft.

Area = 75*75
= 5625 ft^2