among all rectangles with perimeter of 40. Find the the area with largest dimension?

Question

write as fraction in lowest terms

Answer 1

The rectangle with the largest area will always be a square.

40/4=10

A square with side lengths of 10 units yields the largest area.

A=s^2

A=10^2

A=100 square units

A=9*11=99 square units

A=8*12=96 square units

and so on...

 

If you need proof...

P=2l+2w

40=2l+2w

20=l+w

20-w=l

A=lw

A=(20-w)w

A=20w-w^2

this is a parabola opening downward; it has a maximum at its vertex, (h,k)

use h=-b/2a

h=-20/-2=10

 

Answer 2

The largest area of any rectangle with fixed (given) perimeter is always a square, in this case 10 = side length.

 

You can prove this by letting P = 2W + 2L (involving widths & lengths) and also WL = 40.  Solve the latter equation for L & substitute into the equation 40 = W + L...