Question

a rectangular lot is to be fenced off along a highway. if the fence along the highway costs 150.00 per yard. on the other sides P100.00 per yard.find the area of the longest lot that can be fenced off for P10,000.00.

Answers

oobleck
If the highway side has length x, and the ends have length y, then
150x + 100(x+2y) = 10000
so, y = (200-5x)/4
The area is
A = xy = x(200-5x)/4 = 1/4 (200x - 5x^2)
dA/dx = 1/4 (200-10x)
dA/dx = 0 at x=20
and y = 25
Bosnian
Draw a rectangle.

Mark the length of the fence next to the road with x.

Mark the width of the fence with y.

If the fence on the highway costs $150 per yard, on the other sides costs $100 per yard, total cost is:

x ∙ $150 per yard on the highway + x ∙ $100 per yard on the other sides
+ 2 y ∙ $100 = $10,000

You can write equation as:

150 x + 100 x + 200 y = 10,000

250 x + 200 y = 10,000

Divide both sides by 200

1.25 x + y = 50

Subtract 1.25 x to both sides

y = 50 - 1.25 x

y = - 1.25 x + 50

Area:

A = x ∙ y

A = x ∙ ( - 1.25 x + 50 )

A = - 1.25 x ² + 50 x

First derivative:

A'(x) = - 1.25 ∙ 2 x + 50

A'(x) = - 2.5 x + 50

The function has a maximum or minimum at the point where the first derivative is equal to zero.

A'(x) = 0

- 2.5 x + 50 = 0

Add 2.5 x to both sides

50 = 2.5 x

2.5 x = 50

Divide both sides by 2.5

x = 20 yd

Second derivative test:

If f′(x₀) = 0 and f′′(x₀) > 0

function has a local minimum at x₀

If f′(x₀) = 0 and f′′(x₀) < 0

function has a local maximum at x₀

If f′(x₀ ) = 0 and f′′(x₀) = 0

higher order tests must be used

In this case:

A"(x) = [ A'(x) ] ' = ( - 2.5 x + 50 )' = - 2.5 < 0

This means for x = 20 yd function has a maximum.

y = y(x max) = y(20) = - 1.25 x + 50 = - 1.25 ∙ 20 + 50 = - 25 + 50 = 25 yd

A = x ∙ y

Maximum area:

Amax = 20 ∙ 25 = 500 yd²

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