You plan to build four identical rectangular sheep pens in a row. Each adjacent pair of pens share a fence between them. You have a total of 368 feet of fence to use. Find the dimension of each pen such that you can enclose the maximum area.
5w+8L=368
5w-5w+8L=368-5w
8L/8=368/8-5w/8
L=-5/8w+46
this is where I get stuck
Well, it seems like you've made a good start! Now, let's try to find the dimensions of each pen to maximize the enclosed area.
Since we have four pens in a row, we can divide the total fence length by 4 to get the length of each shared fence between adjacent pens.
368 feet of fence / 4 pens = 92 feet of shared fence
Let's call the width of each pen "w" and the length "L". We have two adjacent pens sharing a fence, so the total length of fence used for the length of both pens is 2L. Therefore,
2L = 92 feet
Now let's consider the fences used for the width of each pen. Each pen has two fences on its width, so the total length of all the width fences is 4w. Therefore,
4w = 368 feet - 92 feet = 276 feet
Now we have two equations:
2L = 92
4w = 276
To maximize the area, we need to find the dimensions that will maximize the length "L" while keeping "w" as large as possible.
From the first equation, we can solve for L:
L = 92 / 2 = 46 feet
Now, let's substitute this value of L into the second equation:
4w = 276
w = 276 / 4 = 69 feet
So, the dimensions of each pen to enclose the maximum area are:
Width (w) = 69 feet
Length (L) = 46 feet
Now, we can find the enclosed area by multiplying the width and length:
Area = w * L = 69 feet * 46 feet = 3174 square feet
So, each pen should have a width of 69 feet and a length of 46 feet in order to enclose the maximum area.
To find the dimensions of each pen that would enclose the maximum area, we will need to use the given equation and find the maximum value of the area.
Let's start by expressing the area (A) of each pen in terms of one variable.
Since each pen is rectangular, the area can be calculated as the product of its length (L) and width (w): A = Lw.
We are given the equation 5w + 8L = 368.
Let's solve it for one variable to substitute it into the area equation.
We can solve for L in terms of w by isolating L in the equation:
8L = 368 - 5w
Divide both sides by 8:
L = (368 - 5w) / 8
Now, substitute this expression for L in the area equation:
A = (368 - 5w) / 8 * w
To find the maximum area, we need to find the value of w that maximizes the function A. To do this, we can take the derivative of A with respect to w and set it equal to zero.
dA/dw = (368 - 5w) / 8 - (5/8)w
Set dA/dw equal to zero and solve for w:
(368 - 5w) / 8 - (5/8)w = 0
Multiply through by 8 to clear the fraction:
368 - 5w - 5w = 0
Combine like terms:
368 - 10w = 0
Subtract 368 from both sides:
-10w = -368
Divide both sides by -10 to solve for w:
w = 368 / 10
w = 36.8
Now that we have the value of w, we can substitute it back into the equation for L:
L = (368 - 5w) / 8
L = (368 - 5(36.8)) / 8
L = (368 - 184) / 8
L = 184 / 8
L = 23
Therefore, the dimensions of each pen that would enclose the maximum area are:
Width (w) = 36.8 feet
Length (L) = 23 feet
To find the dimensions of each pen such that you can enclose the maximum area, you should first understand that the area of each rectangular pen is given by the product of its length (L) and width (w), expressed as A = L × w.
You have been given the constraint that each adjacent pair of pens share a fence between them. This means that a total of three fences will be shared between the four pens, leaving only one fence at each end of the row. Therefore, you can conclude that three sides of each pen will have a length equal to w, while the remaining side will have a length equal to 2L (since each pen has 2 adjacent pens).
To determine the dimensions yielding the maximum area, you need to maximize the area function A = L × w while satisfying the given constraint: 5w + 8L = 368.
Let's solve the constraint equation for w:
5w = 368 - 8L
w = (368 - 8L) / 5
Now, substitute the expression for w in the area function:
A = L × [(368 - 8L) / 5]
To maximize the area, you need to find the value of L that maximizes A. This can be done by taking the derivative of A with respect to L and setting it equal to zero:
dA/dL = (368 - 16L) / 5 = 0
Solving for L:
368 - 16L = 0
16L = 368
L = 368 / 16
L = 23
Substitute L = 23 back into the expression for w:
w = (368 - 8(23)) / 5
w = (368 - 184) / 5
w = 184 / 5
w = 36.8
Therefore, by rounding to the nearest whole number, each pen should have dimensions approximately 37 feet by 23 feet to enclose the maximum area.