A fence is to be built to enclose a rectangular area of 310 square feet. The fence along three sides is to be made of material that costs 3 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.

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now i need to buy my materials to fence my yard. using the dimensions above,what would i pay if each foot of fencing costs $12? write an equation and solve

Well, it seems like we have a fencing situation here.

Let's call the length of the rectangular area "x" and the width "y". So, the equation for the area is xy = 310.

Now, let's consider the cost of the fence. We have three sides made of material that costs 3 dollars per foot, which means the total cost of these sides is 3(x + 2y) dollars. The fourth side costs 14 dollars per foot and its length is x, so the cost of this side is 14x dollars.

To find the most economical dimensions, we need to minimize the cost. So, we need to minimize the cost function C(x) = 3(x + 2y) + 14x.

However, we have an equation that connects x and y: xy = 310. To simplify things, let's solve this equation for y: y = 310/x.

Now, substituting this value of y into the cost function, we have C(x) = 3(x + 2(310/x)) + 14x.

To minimize the cost, we can differentiate the cost function with respect to x and set it equal to zero. So, let's do the math... *clown noises*

*differentiation noises*

Now, let's solve the equation C'(x) = 0... *drumroll* ...

*tension rises*

And we get x = √(310/2) ≈ 12.5!

Plugging this value back into the equation xy = 310, we find y ≈ 24.8.

So, the most economical dimensions for the enclosure are approximately 12.5 feet by 24.8 feet.

Now, go forth and build the perfect economical fence! Good luck, my friend!

To find the most economical dimensions for the enclosure, we need to minimize the cost of the fence.

Let's assume the width of the rectangle is w and the length of the rectangle is l.

We are given that the total area enclosed by the fence is 310 square feet, so we have the equation:
w * l = 310.

The cost of the fence can be calculated as follows:
Cost = 3(w + l) + 14(l).

Now let's simplify the equation using the given information and the equation for the cost of the fence:
w * l = 310.
Cost = 3(w + l) + 14l.

To minimize the cost, we can use calculus. Taking the derivative of the cost equation with respect to w, and setting it equal to 0, we can find the critical point where the cost is minimized.

dCost/dw = 3 + 0 - 0 = 0.

This implies that 3 + l = 0.

Simplifying, we find that l = -3.

However, we are seeking real and positive solutions, so l = 3.

Substituting this value of l into the equation w * l = 310, we can solve for w:
w * 3 = 310.
w = 310 / 3.
w ≈ 103.333.

So, the dimensions of the enclosure that are most economical to construct are approximately 103.333 feet by 3 feet.

To find the dimensions of the enclosure that is most economical to construct, we need to minimize the cost of the fence.

Let's assume the length of the rectangular area is L and the width is W.

The area of the rectangle is given as 310 square feet, so we have the equation: L * W = 310.

We also know that there are three sides of the fence, two lengths, and one width, that cost $3 per foot, and one side, which is the remaining width, that costs $14 per foot.

The cost of the fence is given by the equation: Cost = 3 * (2L + W) + 14 * W = 6L + 17W.

Now, we can substitute the value of L from the area equation into the cost equation to get a cost function with only one variable: Cost = 6(310/W) + 17W.

To find the dimensions that minimize the cost, we need to find the value of W that minimizes the cost function.

To do this, we can take the derivative of the cost function with respect to W, set it equal to zero, and solve for W.

d(Cost)/dW = -18620/W^2 + 17.

Setting this derivative equal to zero gives us -18620/W^2 + 17 = 0.

Rearranging this equation, we have 17 = 18620/W^2.

Simplifying further, we get W^2 = 18620/17.

Taking the square root of both sides, we find W = sqrt(18620/17) ≈ 20.62.

Now, substituting this value of W back into the area equation, we can solve for L: L = 310/W ≈ 15.02.

Therefore, the dimensions of the enclosure that are most economical to construct are approximately Length (L) ≈ 15.02 feet and Width (W) ≈ 20.62 feet.