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.

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

Well, well, well! Looks like we have a fencing conundrum! Let's solve it with a clownish twist, shall we?

First, let's label our sides. We've got the highway side, which costs P150.00 per yard to fence, and the other three sides, which cost P100.00 per yard. Let's call the length of the highway side "x" and the other two sides "y."

Now, we know that the total cost of fencing should not exceed P10,000.00. So, let's start by considering the cost of the highway side: P150.00 multiplied by the length "x."

Our equation for the cost of the highway side is P150.00 * x ≤ P10,000.00.

Next, we have the cost of the other two sides, which is P100.00 per yard. Since there are two of them, the total cost of the non-highway sides is P100.00 * 2 * y.

Our equation for the cost of the non-highway sides is P100.00 * 2 * y ≤ P10,000.00.

Now, since we're looking for the area of the longest lot that can be fenced off, we want to maximize the area. The area of a rectangular lot is given by the formula A = x * y.

To proceed, we need to get rid of those pesky inequalities. Let's solve each one individually:

P150.00 * x ≤ P10,000.00
x ≤ P10,000.00 / P150.00
x ≤ 66.67 yards (approximately)

P100.00 * 2 * y ≤ P10,000.00
y ≤ P10,000.00 / (P100.00 * 2)
y ≤ 50 yards

Great! Now that we have our maximum values for x and y, let's calculate the area:

A = x * y
A = 66.67 yards * 50 yards
A = 3,333.33 square yards

So, there you have it! The area of the longest lot that can be fenced off for P10,000.00 is approximately 3,333.33 square yards. Enjoy your clownishly spacious lot!

To find the area of the longest lot that can be fenced off for P10,000.00, we need to optimize the dimensions of the lot.

Let's assume the width of the lot is w yards, and the length of the lot (parallel to the highway) is l yards.

The cost of the fence along the highway is P150.00 per yard, and the cost of the fence on the other sides is P100.00 per yard.

The cost of the fence along the highway will be 150 * l, and the cost of the fence on the other sides will be 100 * (2w + 2l).

According to the problem, the total cost of the fence should be P10,000.00. Therefore, we can set up the equation:

150 * l + 100 * (2w + 2l) = 10,000

Simplifying the equation, we get:

150l + 200w + 200l = 10,000
350l + 200w = 10,000

To solve this equation, we need to express one variable in terms of the other. Let's solve for l:

350l = 10,000 - 200w
l = (10,000 - 200w) / 350

To maximize the area, we need to find the maximum value for l, which means minimizing w. Since w cannot be negative (as it represents a length), we set w to zero and calculate l:

l = (10,000 - 200 * 0) / 350
l = 10,000 / 350
l ≈ 28.57 yards

Now we know that l ≈ 28.57 yards. To find the corresponding width, we substitute this value back into the equation:

350l + 200w = 10,000
350 * 28.57 + 200w = 10,000
10,000 + 200w = 10,000
200w = 10,000 - 10,000
200w = 0
w = 0

Therefore, the maximum area for P10,000.00 is achieved when the width of the lot is 0 yards. This implies that the lot is a line segment along the highway with a length of approximately 28.57 yards.

So, the area of the longest lot that can be fenced off for P10,000.00 is effectively zero.

To find the area of the longest lot that can be fenced off for P10,000.00, we need to consider the cost of fencing along the highway and the cost of fencing on the other sides.

Let's assume the length of the lot is L and the width of the lot is W. So, the cost of fencing along the highway would be 150.00 per yard multiplied by the length L. The cost of fencing on the other sides would be 100.00 per yard multiplied by 2 times the sum of the length L and the width W.

The total cost of fencing can be calculated as follows:

Total Cost = Cost of Fencing along the Highway + Cost of Fencing on the other Sides

Now, let's set up the equation:

10000 = (150 * L) + (100 * 2 * (L + W))

Simplifying the equation, we get:

10000 = 150L + 200(L + W)

10000 = 150L + 200L + 200W

Combining like terms:

10000 = 350L + 200W

Rearranging the equation:

350L + 200W = 10000

Now, we need to express one variable in terms of the other to find the maximum area. Let's solve the equation for L in terms of W:

350L = 10000 - 200W

L = (10000 - 200W) / 350

To maximize the area, we need to substitute the value of L back into the equation for the area:

Area = L * W

Area = [(10000 - 200W) / 350] * W

Now, we have the area equation in terms of just one variable W. We can find the maximum area by finding the value of W that maximizes this area equation.

To do this, we can take the derivative of the area equation with respect to W, set it equal to zero, and solve for W to find the critical point. Then, we can substitute this value back into the area equation to find the maximum area.

Differentiating the area equation:

d(Area) / dW = (10000 - 200W) / 350 - (200W) / 350

Setting the derivative equal to zero:

(10000 - 200W) / 350 - (200W) / 350 = 0

Simplifying the equation:

10000 - 200W - 200W = 0

10000 - 400W = 0

-400W = -10000

W = -10000 / -400

W = 25

Now, substitute the value of W back into the area equation:

Area = [(10000 - 200W) / 350] * W

Area = [(10000 - 200(25)) / 350] * 25

Area = [(10000 - 5000) / 350] * 25

Area = (5000 / 350) * 25

Area = 14.2857 * 25

Area = 357.14 square yards

Therefore, the area of the longest lot that can be fenced off for P10,000.00 is 357.14 square yards.

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