Find the dimensions of a rectangle with perimeter 60 m whose area is as large as possible.

**Answer 1**

Width : w

Length : ℓ

Perimeter : P

Area : A

P = 60

2(w + ℓ) = 60

w + ℓ = 30

w = 30 – ℓ

A = ℓ * w

A = ℓ * (30 – ℓ)

A = 30ℓ – ℓ²

You can see that A is a function of ℓ.

A = f(ℓ) = 30ℓ – ℓ²

A will be maximum when the dericate of A will be null

A’ = f'(ℓ) = 30 – 2ℓ

A’ = 0

30 – 2ℓ = 0

2ℓ = 30

ℓ = 15

w = 30 – ℓ = 30 – 15 = 15

The area will be maximum when : w = ℓ = 15, so when the perimeter is 60 m, and when you’ve got a square

A = ℓ * w = 15 * 15 = 225 m²

**Answer 2**

enable x = length of the rectangle y = width of the rectangle Perimeter = P = one hundred twenty = 2(x + y) 60 = x + y or y = 60 – x section = A = xy considering that y = 60 – x A = x(60 – x) A = 60x – x^2 Differentiating the above function, (dA/dx) = 60 – 2x and to ensure the length which will maximize the realm, set (dA/dx) = 0. for this reason (dA/dx) = 0 = 60 – 2x and fixing for “x”, x = 60/2 x =30 meters to ensure “y”, y = 60 – x y = 60 – 30 y = 30 answer: Dimensions of the rectangle for optimum section (given the circumstances of the placement) will be 30 m x 30 m. that’s actually a sq..

**Answer 3**

my first guess is:

P=60 => p=30

l+L = 30

So, for A=l*L to be the greatest and it won’t be a square, l=14 and L=16 or viceversa.

I’m not entirely sure about it though

**Answer 4**

A=xy

2x+2y=60

2y=(60-2x)

y=(60-2x)/2

y=(30-x)

A=x(30-x)

A=30x-x^2

A ‘ (x)=30-2x

A'(x)=0

0=30-2x

2x=30

x=15

2x+2y=60

2.15+2y=60

2y=30

y=15