Question

Of all the rectangles each of which has perimeter 40 metres find one which has maximum area.Find the area also.

Answer 1

Answer:

Maximized area when the rectangle is a square with side 10 m, area 100 m².

Hope this helps you.

Step-by-step explanation:

By the principle of insufficient reason, we would expect the answer to be the most symmetric candidate, namely a square.  Its side would then be 10 m and its area 100 m².

Now to be certain..

Let x and y be the length and width of the rectangle.  Then...

perimeter = 40 m

=> 2 ( x + y ) = 40

=> x + y = 20

=> y = 20 - x

The area is then given by

A = xy = x ( 20 - x ).

We need to maximize this.  How you do this depends on what you're learning in class.

Option A

If you're learning about quadratics and parabolas, notice that A is a quadratic in x.  Plotting the curve A = x ( 20 - x ) = -x² + 20x, this is a parabola with a maximum at x = 20/2 = 10.  So the value of A (the area) is a maximum when x = 10.

This gives y = 20 - x = 20 - 10 = 10 (so it's a square) and area A = xy = 100.

Option B

You're learning about calculus.

To maximize, find where dA/dx = 0.

dA/dx = -2x + 20 = 0  =>  x = 10.

Since we know that 0 ≤ x ≤ 20, the "critical points" where a maximum can occur are at 10 and at the end points 0 and 20.  Just check the value of A at each of these  ( A(0)=0, A(10)=100, A(20)=0 ) to see which is greatest.

Option C

You're learning about inequalities.

Complete the square and use the fact that u² ≥ 0, with equality when u = 0.

So here, A = - ( x² - 20x ) = - ( x - 10 )² + 100 ≤ 0 + 100 = 100.  The upper bound of 100 is attained when (x-10)² = 0, and this is when x = 10.