SPs Using Derivatives?

Question

Find the dimensions of a rectangle with perimeter 100m whose area is as large as possible.

I wrestled with this one for a while..and I think I did it right! I just need confirmation! THanx!

Answer 1

Let x, y denote the sides of the rectangle. Then we know that

2y + 2x = 100, ie. y = 50 - x.

Thus the area is given by

A(x) = x(50 - x) = 50x - x^2.

This is a quadratic with negative leading term and so has a maximum at A'(x) = 0. Thus, we differentiate A and find the root of the derivative:

A'(x) = 50 - 2x = 0 iff x = 25.

Thus x = 25 and y = 25, and so it's a square of side 25 m.

Answer 2

You're right the rectangle that produces the max area is a square.

The general case for a rectangle with fixed perimeter is

P = 2L + 2W; solving for W you get W = (P - 2L)/2 so the area, as a function of the single variable L is

f(L) = LW = L [P-2L]/2 = [LP - 2L^2] / 2

Find critical value(s) for first derivative with respect to L:

[P - 4L] / 2 = 0 => L = P/4 which when substituted back (line 3)means

W also = P/4 [If L and W are the same dimension in a rectangle,

you have a square.]

[Verification you have a relative max for function...take second derivative and it is negative: f'(L) = -2.

Answer 3

Area=xy and x+y=50

Area=x(50-x)=50x-x^2

d(Area)/dx=0 or 50-2x=0
x=25.

Answer 4

If I understend the question correctly it is rectangle 25x25 m - it is square

Answer 5

It is true only when the rect. is a square. Such a simple thing to do.