let’s say you have 4x feet of fencing, where x is any number. What, in terms of x, should the dimensions of the largest rectangular enclosure be, whose perimeter is 4x feet? What would the area be?
To find the dimensions and area of the largest rectangular enclosure with a perimeter of 4x feet, we need to use some algebra.
Let's assume the length of the enclosure is L and the width is W. The perimeter of a rectangle is calculated by adding the lengths of all its sides. In this case, it is given as 4x feet.
Perimeter = 2(L + W)
Since we are given that the perimeter is 4x, we can substitute this value into the equation:
4x = 2(L + W)
Now let's solve this equation for either L or W in terms of x. Let's solve it for L:
2L = 4x - 2W
Divide both sides by 2:
L = 2x - W
Now we have an expression for the length in terms of x and the width. But since we want to find the largest rectangular enclosure, we need to maximize the area. The area of a rectangle is calculated by multiplying its length and width:
Area = L * W
Substitute the expression for L:
Area = (2x - W) * W
Now we have the area in terms of W. To find the maximum area, we need to find the value of W that maximizes this expression.
One way to do this is by finding the vertex of the quadratic equation (the area equation is quadratic in W). The vertex will give us the maximum value.
The quadratic equation is:
Area = 2xW - W²
The vertex of a quadratic equation in the form Ax² + Bx + C can be found using the formula:
x = -B / (2A)
In our case, A = -1, B = 2x, and C = 0. Therefore:
W = -2x / (2*(-1))
Simplifying further:
W = x
So we find that the value of W that maximizes the area is x. Now we substitute this value back into the expression for L:
L = 2x - x
Simplifying further:
L = x
Therefore, the dimensions of the largest rectangular enclosure with a perimeter of 4x feet are x by x, or simply x by x. The area is calculated as L * W:
Area = x * x
The area of the largest rectangular enclosure is x² square feet.