Question

The perimeter of a rectangle is 38m. Find the dimensions of the rectangle that will contain the greatest area.

Answer 1

can you help me?

Answer 2

Let the sides of the rectangle be x and y

2x+2y=38
x+y=19
Area: xy = x(19-x)
AM/GM inequality
x(19-x) <= (x+19-x)²/4 = 19²/4
x=y=19/2 gives max area 361/4 and occurs when rectangle is a square

Answer 3

another method..

let x be the width and y length of the rectangle
P=2(x+y)
from the given info p=38
2(x+y)=38 (divide both sides by 2)
x+y=19 (subtract x from both sides)
y=19-x

area=x*y
and we want
x*y=max
from the above calculation y=19-x
x*y=x*(19-x)=max
\[x(19-x)=19x-x ^{2}=\max\]
by completing the square method
\[19x-x ^{2}=-(x+\frac{ 19 }{ 2 })+\frac{ 19 }{ 2 }\]
now draw the graph -(x+\frac{ 19 }{ 2 })+\frac{ 19 }{ 2 }\]

or simply the vertex gives the maximum values of x and y.
\[x=\frac{ 19 }{ 2 } \] and
\[y=\frac{ 19 }{ 2 } \]