Answer to Question #325457 in Calculus for Tina

Question

1. Find the dimension of a rectangle with perimeter 200 m and whose area is as large as possible

2. The sum of two positive numbers is 10. Find the numbers if the sum of their squares is a minimum

Answer 1

1. Let the length of the rectangle be x m.

Then width becomes (200 - 2x)/2 = (100 - x) m.

Let the area be y = length times width = x(100-x) sq m.

A plot of y = 100x - x²

shows y (i.e. area) is max when x = 50 m.

Dimensions for maximum area is 50 x 50 meters.

Area is 2,500 m².

2. If the sum of two positive numbers, x and y is 10

x + y = 10

y = 10 - x

The sum of their squares is$x^2 + (10 - x)^2 = x^2 + 100 - 20x + x^2 = 2x^2 - 20x + 100$

The minimum will occur when the derivative = 0

f’(x) = 4x - 20

4x - 20 = 0

To get the value of x,

4x = 20

4x/4 = 20/4

x = 5

That is the minimum occurs at (x,y) = (5,5)

and the minimum possible value of the sum of the squares is

5²+ 5² = 25 + 25 = 50