A farmer decides to enclose a rectangular garden, using the side of a barn as one side of the rectangle. What is the maximum area that the farmer can enclose with 80 ft of fence? what should the dimensions of the garden be to give this area.
The maximum area that the farmer can enclose with 80 ft of fence is ___sq ft.
The dimensions of the garden to give this area is 40 ft by ____ft
Normally the largest area is enclosed by a square but this is only when the perimeter is 4 sides. In this problem the barn is the forth side with the 80 ft of fence used for only three sides. The complication is that the barn can contribute any lenght for its side. So the barn length as a variable turns the problem into a second order equation or the quadratic.
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The equations are 2x+y=80 and 2x=y. "x" being the widthof the fence coming off the barn and "y" being the length connecting those two pieces. When one side is as big as nessasary (aka the barn, river, wall, etc.) the length is double the width. The perimeter is usually defined =2x+2y, but in this case one y is already there so the equation 2x+y=80 (and 2x+y=40)works. Try that out :)
To find the maximum area that can be enclosed with 80 ft of fence, we can use the formula for the perimeter of a rectangle, which is P = 2L + 2W, where P is the perimeter, L is the length, and W is the width of the rectangle. Since the farmer wants to use the side of a barn as one side of the rectangle, we can set the equation as follows:
80 ft = L + 2W
Next, we can express L in terms of W by subtracting 2W from both sides of the equation:
L = 80 ft - 2W
To find the area of the rectangle, we use the formula A = L * W. Since we have an expression for L in terms of W, we can substitute it into the area formula:
A = (80 ft - 2W) * W
Expanding and reordering the equation, we get:
A = 80W - 2W^2
To find the maximum area, we need to maximize this equation. Since it is in the form of a quadratic function, we know that the maximum occurs at the vertex of the parabola. The formula to find the x-coordinate of the vertex of a quadratic function in the form of f(x) = ax^2 + bx + c is x = -b / (2a). In our case, a = -2 and b = 80, so we can find the value of W at the vertex as follows:
W = -80 / (2 * -2) = 20 ft
Now that we have the width, we can substitute it back into the equation for L to find the length:
L = 80 ft - 2(20 ft) = 40 ft
Therefore, the maximum area that the farmer can enclose with 80 ft of fence is 40 ft by 20 ft, which is 40 * 20 = 800 sq ft.