A rectangular lot is bordered on one side by a stream and on the other three sides by 600m of fencing. Find the dimensions of the lot if area is a maximum.
150x300
May i ask how you found it?
To find the dimensions of the lot that will maximize the area, we need to determine the relationship between the dimensions and come up with an equation for the area.
Let's assume the length of the lot is L and the width is W.
Given that the lot is bordered on one side by a stream, the length of the lot will be equal to the length of fencing on the two vertical sides, while the width of the lot will be equal to the fencing on the top and bottom sides.
From the given information, we can deduce the following equation based on the perimeter:
2L + W = 600
Now, let's express the area of the lot in terms of L and W. The formula for the area of a rectangle is A = L × W.
Substituting the value of W from the perimeter equation, we have:
A = L × (600 - 2L)
To find the dimensions that maximize the area, we need to find the critical points of this equation. We can do this by taking the derivative of the equation with respect to L and setting it equal to zero.
dA/dL = 600 - 4L = 0
Solving this equation, we find L = 150. Substituting this back into the perimeter equation, we find W = 150.
Therefore, the dimensions of the lot that maximize the area are 150m × 150m.
you might look at the many similar questions below. I'll give you the shortcut, which always works in this kind of situation.
The maximum area is achieved when the fencing is divided equally among the lengths and the widths. In this case, we have one length and two widths. Half of 600 is 300, so there is one length of 300, and two widths of 150 each.
This also works where the area is divided into a grid of equal-size pens, each of width x and length y.
If there are m widths of x, and n lengths of y, then if the fencing has length z,
mx+ny = z
The area is mn(xy)
= mn(x)(z-mx)/n
= mx(z-mx)
= mzx - m^2 x^2
This is just a parabola, with vertex at x = mz/2m^2 = z/2m
That is, the m widths add up to z/2, or half the fencing.