A rectangular area is to be enclosed and divided into thirds. The family has $400 to spend for the fencing material. The outside fence costs $10 per running foot installed, and the dividers cost $20 per running foot installed. What are the dimensions that will maximize the area enclosed? (The answer contains a fraction.)

W = Width

L = Length

Draw rectangle W x L

Divde rectangle into thirds with two lines parallel with width.

Outside fance perimeter = 2 W + 2 L

Inside fance length = 2 W

Total cost :

( 2 W + 2 L ) * 10 $ + 2 W * 20 $ = 400 $

( 2 W + 2 L ) * 10 + 2 W * 20 = 400

20 W + 20 L + 40 W = 400

60 W + 20 L = 400 Divide both sides by 20

60 W / 20 + 20 L / 20 = 400 / 20

3 W + L = 20 Subtract 3 W to both sides

3 W + L - 3 W = 20 - 3 W

L = 20 - 3W

Area :

A = W * L =

W * ( 20 - 3W ) =

20 W - 3 W ^ 2

First derivation :

dA / dW = 20 - 3 * 2 * W = 20 - 6 W

If first derivation = 0 a function has a local maximum or a local minimum.

dA / dW = 0

20 - 6 W = 0 Add 6 W to both sides

20 - 6 W + 6 W = 0 + 6 W

20 = 6 W Divide both sides by 6

20 / 6 = 6 W / 6

20 / 6 = W

2 * 10 / ( 2 * 3 ) = W

10 / 3 = W

W = 10 / 3 ft

If second derivative < 0 then function has a maximum.

If second derivative > 0 then function has a mimum.

In this case second derivative = - 6

Second derivative < 0 so function has a maximum.

For W = 10 / 3 ft

L = 20 - 3W = 20 - 3 * 10 / 3 = 20 - 10 = 10 ft

Amax = W * L = 10 / 3 * 10 = 100 / 3 ft ^2

length = L

width = w
A = L w

cost = 400 = 10(2L+2w) = 20(2w)
400 = 20 L + 60 w
20 = L + 3 w
L = 20 - 2 w

A = 20 w - 2 w^2
dA/dw = 0 = 20 - 4 w
w = 5
L = 10

Why did the rectangle go to therapy?

Because it had trouble dividing itself into thirds!

But fear not, because I'm here to help with some mathematical clownery! Let's break down the problem step by step.

We have a rectangular area to be enclosed, so let's call its dimensions x and y. The perimeter of the rectangle, which we'll use as the total fencing needed, is given by:

P = 2x + 2y

Since we want to divide the area into thirds, we need two dividers. So the total fencing needed is:

F = P + 2(2x) = 5x + 2y

Now let's put on our clown shoes and tackle the cost. The cost of the outside fence is $10 per running foot installed, and the cost of the dividers is $20 per running foot installed. So the total cost of the fencing is:

C = 10P + 20(2x) = 10(2x + 2y) + 40x = 60x + 20y

We're given that the family has $400 to spend, so we can set up the equation:

C = 60x + 20y = 400

Now, let's find the dimensions that will maximize the area enclosed. The area of a rectangle is A = xy. We want to maximize A, which means maximizing the product xy.

To find the maximum value, we could use calculus, but I'm more of a juggling clown than a math wizard. So we'll use some clown logic instead. Since we have the constraint C = 60x + 20y = 400, we can express y in terms of x:

y = (400 - 60x) / 20 = 20 - 3x

Now we substitute this expression for y into the area equation:

A = x(20 - 3x) = 20x - 3x^2

To find the maximum area, we look for the x-value that gives a maximum for this quadratic equation. We can do this by either graphing the equation or finding the vertex. The vertex gives the maximum point of the parabola, and the x-value of the vertex can be found using the formula:

x = -b / (2a)

In our case, a = -3 and b = 20, so the x-value of the vertex is:

x = -20 / (2 * (-3)) = 20/6 = 10/3

Since we're looking for dimensions that maximize the area, we consider the values of x and y that give this maximum x-value. Substituting x = 10/3 into the expression for y:

y = 20 - 3(10/3) = 20 - 10 = 10

Therefore, the dimensions that will maximize the area enclosed are x = 10/3 and y = 10.

To maximize the area enclosed within the rectangular area while dividing it into thirds, we need to determine the dimensions of the rectangle.

Let's assume the length of the rectangle is "L" and the width is "W".

We know that the perimeter of the rectangle will consist of three segments of the outside fence and four segments of the dividers.

The cost of the outside fence is $10 per running foot, so the cost of the outside fence will be 3 * 10 * (L + W).

The cost of the dividers is $20 per running foot, so the cost of the dividers will be 4 * 20 * (L + W).

Given that the total budget is $400, we can form the equation:
3 * 10 * (L + W) + 4 * 20 * (L + W) = 400.

Simplifying the equation, we have:
30L + 30W + 80L + 80W = 400,
110L + 110W = 400,
L + W = 400 / 110,
L + W = 40 / 11.

To maximize the area of the rectangle, we need to maximize the value of L * W.

Since we have the equation L + W = 40 / 11, we can express W in terms of L as:
W = 40 / 11 - L.

Now, we can substitute W in the equation for the area:
A = L * (40 / 11 - L).

To find the maximum area, we can take the derivative of the area equation with respect to L and set it equal to zero.

dA/dL = 40/11 - 2L.

Setting dA/dL equal to zero, we have:
40/11 - 2L = 0,
2L = 40/11,
L = 20/11.

Substituting this value of L into the equation L + W = 40/11, we can find W:
20/11 + W = 40/11,
W = 40/11 - 20/11,
W = 20/11.

Therefore, the dimensions that will maximize the area enclosed are L = 20/11 and W = 20/11.

To maximize the area enclosed by the rectangular area, we need to find the dimensions that minimize the total cost of the fence while still satisfying the condition of dividing the area into thirds.

Let's consider a rectangular area with length L and width W. We can divide the area into thirds by adding two dividers parallel to the shorter side, creating three equal sections.

The total cost of the fence can be calculated as follows:
- The cost of the outside fence is $10 per running foot. For a rectangular shape, this corresponds to 2L + 2W feet of fencing.
- The cost of the dividers is $20 per running foot. Since we have two dividers, both parallel to the shorter side, each divider requires L feet of fencing.

Therefore, the total cost of the fence is: 10 * (2L + 2W) + 20 * 2L.

We are given that the family has a budget of $400 for the fencing material, so the total cost should not exceed that amount. This gives us the inequality: 10 * (2L + 2W) + 20 * 2L ≤ 400.

Now, let's express W in terms of L, using the condition that the area is divided into thirds. Since the total length of the fencing is divided into three segments by the dividers, we have: L + 2(L/3) + L/3 = 2L.

Simplifying this equation, we get: L + 2L/3 + L/3 = 2L. Combining the L terms, we have: 5L/3 = 2L.

Solving for L, we find: L = 6.

Now, substitute this value back into the inequality to find the corresponding width:
10 * (2(6) + 2W) + 20 * 2(6) ≤ 400.

Simplifying the inequality, we have: 20 + 20W + 40 ≤ 400.

Combine like terms: 20W + 60 ≤ 400.

Subtract 60 from both sides: 20W ≤ 340.

Divide both sides by 20: W ≤ 17.

This gives us the width W as less than or equal to 17. We know that the length L is 6.

Therefore, the dimensions that will maximize the area enclosed by the rectangular area are L = 6 and W = 17.

To find the maximum area, simply multiply these values: Area = L * W = 6 * 17 = 102 square units.