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.

Answer 1

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

Answer 2

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