How do you find the maximum area of a rectangle with perimeter of 40 ft? I need to know how to find Max. Area?

Question

So how the HELL do i find it??!?!?! lol, that didnt really help

Answer 1

let the length of the rectangle be x
width will be =(40/2)-x=20-x
area=x(20-x)
=-(x^2-20x)
-(x^2-20x+100)+100
=>100-(x-10)^2
for maximum
(x-10)^2 must be 0
or x=10
so it must be a square with side 10 ft.

if you take the calculusroute
A=x(20-x)
=20x-x^2
dA/dx=20-2x
for max set this to zero
so 20-2x=0
and x=10
d^2A/dx^2=-2 negative
so thevalue is maximum for x=10

Answer 2

A rectangle has a length and a width. Let's denote L to be the length of the rectangle, and W to be the width. Then, the perimeter of a rectangle is given as

P = length of each of the 4 sides.

But, we have a length and a width. The opposite sides of a rectangle have the SAME size. Two of the opposite sides are of length L, and two of the opposite sides of are of width W. Therefore,

P = L + L + W + W
P = 2L + 2W

We're given that the perimeter P = 40 though. Therefore,

40 = 2L + 2W

We can divide all of this by 2, to get

20 = L + W

The area is what we want to maximize. We have a formula for the area of the rectangle.

A = LW

However, we just solved that 20 = L + W, so W = 20 - L.
We plug W = 20 - L into the area formula, to get

A = L(20 - L), or
A = 20L - L^2

So now, since we have one variable, we can declare this to be our area function, A(L).

A(L) = 20L - L^2

In order to find the maximum area, we need to find the derivative, and then make it 0.

A'(L) = 20 - 2L

Now, we make A'(L) = 0

0 = 20 - 2L
2L = 20
L = 10

So L = 10 is *where* the maximum area occurs. To actually CALCULATE the maximum area, we plug in L = 10 into our function A(L).

A(10) = 20(10) - (10)^2 = 200 - 100 = 100

So the maximum area is 100 square feet.

Answer 3

Well, I have to dispute Robin's assessment - this problem is not THAT complicated...

Since the perimeter of the rectangle is 40, if we call the length of two opposite sides x, then the length of the other two sides is (20-x). So, the area of the rectange is f(x) = x(20 - x) = 20x - x^2. That's an upside-down parabola, so now just find the point where it hits its maximum value. You'll get x = 10 for that, therefore the area is 10(20 - 10) = 100

Answer 4

Well the easy answer to this is that a rectangle has a maximum area when it is a square - therefore the sides would be 10ft eact and area = 100ft squared.

The complicated answer would involve setting out and solving a quadratic equation for a maximum value. Start by writing Area = base * height, Perimiter = 2 * Base + 2 * height and go from there

Answer 5

we have S = a*b, a + b =P/2 = 20
according to Cauchy inequality for two positive numbers, we have
a*b <= (a+b)^2 / 4 <=> a*b <= 100
the sign "=" happens when and only when a = b = 10

So the area of the rectangle reaches max value when and only when it is a square with 10ft as the the length of the side, the area will be 100 squared ft

Answer 6

100