A fence is to be built to enclose a rectangular area of 320 square feet. The fence along three sides is to be made of material that costs 6 dollars per foot, and the material for the fourth side costs 14 dollars per foot. Find the dimensions of the enclosure that is most economical to construct.
Ah, the age-old quest for the most economical fence! Let me put my clown nose on and calculate this for you in the funniest way possible.
Let's say the length of the rectangular area is "x" feet, and the width is "y" feet. We want to find the dimensions that minimize our cost.
We know the fence along three sides (2 lengths + 1 width) costs 6 dollars per foot, and the fourth side (remaining width) costs 14 dollars per foot. With me so far? Good!
So, the cost of the three sides is 3 * (6x + 6y) dollars, and the cost of the remaining side is 14y dollars. The total cost, C, is just the sum of these costs:
C = 3(6x + 6y) + 14y
Now, we want to find the minimum cost. To do that, we need to take the derivative of C with respect to either x or y. But since this is a clown show, let's take the derivative with respect to both, just for laughs!
dC/dx = 18
dC/dy = 18 + 14
Setting both of these derivatives equal to zero (because we want the minimum), we get:
18 = 0 and 18 + 14 = 0
Wait, that makes no sense! Silly me, I made a mistake in the calculations. Let me try it again without the clown hat on this time.
Actually, let's think about it logically. The cost per foot for the fourth side is higher, so it would make sense to make the fourth side as short as possible.
Since we know the area (320 square feet), we can express y in terms of x: y = 320/x
Now, let's plug this into our cost equation:
C = 3(6x + 6(320/x)) + 14(320/x)
Simplifying further, we get:
C = 18x + 5760/x + 4480/x
Combining like terms, we have:
C = 18x + 10240/x
To find the minimum cost, we need to find the critical points. But since we want to have some fun, let's skip the calculus and use logic again. The minimum cost occurs when C is as small as possible, which means x should be as small as possible.
So, the dimensions that are most economical to construct would be a square enclosure, with each side being sqrt(320) feet.
I hope you had a laugh with me on this mathematical adventure! Just remember, even clowns can do math (most of the time).
To find the dimensions of the enclosure that is most economical to construct, we can first set up an equation to represent the cost of the fence. Let's assume the length of the rectangular area is L and the width is W.
The cost of the fence along three sides is given by:
3L + W
The cost of the fourth side is given by:
14W
The total cost of the fence is the sum of the costs along three sides and the cost of the fourth side:
Cost = 3L + W + 14W
We also know that the area enclosed by the fence is 320 square feet:
Area = L * W = 320
Now we can solve for the dimensions by finding the critical points of the cost function.
1. Express the cost function in terms of one variable:
Cost = 3L + W + 14W
Cost = 14W + (3L + W)
Cost = 14W + 3L + W
Cost = 15W + 3L
2. Express the area in terms of one variable:
320 = L * W
L = 320 / W
3. Substitute L in the cost equation:
Cost = 15W + 3(320 / W)
4. Differentiate the cost function with respect to W:
dCost/dW = 15 - 960 / W^2
5. Set the derivative equal to zero and solve for W:
15 - 960 / W^2 = 0
15W^2 - 960 = 0
W^2 = 960 / 15
W^2 = 64
W = 8 or W = -8 (discarding negative value)
6. Substitute the value of W back into the equation for L:
L = 320 / W
For W = 8: L = 320 / 8 = 40
Therefore, the dimensions of the enclosure that is most economical to construct are L = 40 ft and W = 8 ft.
To find the dimensions of the enclosure that is most economical to construct, we'll need to consider the costs associated with each side of the fence. Let's start by assigning variables to the dimensions of the rectangular area.
Let's say x is the length of the rectangular area, and y is the width. Since the area is given as 320 square feet, this means xy = 320.
We know that the fence has three sides with a cost of 6 dollars per foot, and one side with a cost of 14 dollars per foot. The total cost of the fence can be expressed as:
Cost = 3(6x) + 14y
Simplifying this equation, we have:
Cost = 18x + 14y
To find the dimensions that minimize the cost, we can use either calculus or algebra. Let's use algebra:
We have xy = 320, so we can express y in terms of x:
y = 320 / x
Now, substitute this expression for y into the cost equation:
Cost = 18x + 14(320 / x)
To minimize the cost, we need to find the values of x that make the derivative of the cost function equal to zero. Differentiating the cost function with respect to x, we get:
dCost/dx = 18 - (4480 / x^2)
Setting this derivative equal to zero and solving for x:
18 - (4480 / x^2) = 0
Multiplying through by x^2, we have:
18x^2 - 4480 = 0
Dividing through by 2, we get:
9x^2 - 2240 = 0
Simplifying further:
9x^2 = 2240
x^2 = 2240 / 9
x ≈ √(2240 / 9)
x ≈ 15.442
Now, substitute this value of x back into the equation for y:
y = 320 / x
y = 320 / 15.442
y ≈ 20.706
Therefore, the dimensions of the enclosure that is most economical to construct are approximately 15.442 feet by 20.706 feet.
If the cheap sides are x, and the expensive side is y, then
xy = 320
c(x) = 6*2x + 6y + 14y = 12x+20y
= 12x + 20(320/x)
c'(x) = 12 - 6400/x^2
minimum cost is at x = 80/√12 ≈ 23
So, the area is roughly 23 by 14
(That's 322 ft^2)