Maximum and minimum problem?

Question

the perimeter of a rectangle is 12 m. Find the dimensions for which the diagonal is as short as possible.

Answer 1

The perimeter of a rectangle is calculated using the following formula:

P = 2L + 2W

But we know that P = 12, so

12 = 2L + 2W. Dividing both sides by 2, we have

6 = L + W

The diagonal of a rectangle D is measured by the Pythagorean relationship

D^2 = L^2 + W^2

But, since 6 = L + W, then 6 - W = L, so

D^2 = (6 - W)^2 + W^2

This means

D = sqrt [(6 - W)^2 + W^2]

So this is the function we want to minimize, which we will now call D(W)

D(W) = sqrt [(6 - W)^2 + W^2]

Since we're eventually going to make D'(W) equal to 0 anyway, let's square both sides, and then differentiate implicitly.

[D(W)]^2 = (6 - W)^2 + W^2

2[D(W)] D'(W) = 2(6 - W) (-1) + 2W

Setting D'(W) to 0, we have

0 = 2(6 - W)(-1) + 2W
0 = -2(6 - W) + 2W
0 = -12 + 2W + 2W
0 = -12 + 4W
-4W = -12, so
W = 3

We get a minimum diagonal at W = 3.
We can get L too, since L = 6 - W = 6 - 3 = 3

Therefore, the dimensions that will obtain the shortest diagonal will be 3 x 3.

Answer 2

Get an expression for the diagonal, and use the derivative to find out when it reaches a minimum. Here's how:

If L and W are the length and width of a rectangle, then its perimeter is 2L+2W. So if 2L+2W=12 then L+W=6. Using the Pythagorean Theorem, the diagonal is √(L² + W²). Leting L=6-W, this can be rewritten as √((6-W)² + W²) = √(36 -12W + W² + W²) = √(2W² -12W+36)

If we were to graph this value against varying values of W, we'd get some kind of curve. If there's a lowest point on this curve, then this low point has to be at the bottom of a "hill" in the curve. This happens when the derivative (slope of the tangent) is zero. The derivative of the diagonal function we made is:

(1/2)(2W² -12W+36)^(-1/2) * (4W - 12) =
(4W - 12) / 2√(2W² -12W+36)

This can only be zero when the top is 0. So it's when W=3. Remember L+W=6, so this means the dimensions of the rectangle with a perimeter 12m and the smallest diagonal are 3m and 3m.

On a side note, you can let "P" be the paramenter to show that a rectangle of a fixed parameters has the biggest diagonal when it's a square.

Answer 3

When it is a square = when all sides are equal = 12m/4 = 3 m each side