Question

The area of a rectangular goat pen is to be 100m². If the length of one side is xmetres, show that the perimeter is (2x+200/2) metres. Prove also that the least perimeter of the pen is 40m.​

Answer 1

Area of rectangle with sides a and b is ab.

We have one side x and area 100 m².

Therefore the second side is:

• 100/x

The perimeter is:

• P = 2(a + b)
• P = 2(x + 100/x) = 2x + 200/x

Proved

Answer 2

Step-by-step explanation:

Given:

Area of rectangular pen = 100 m²

Lenth of one side of pen (a) = x m

To prove:

Perimeter (P) = 2x + 200/x

Least perimeter = 40 m

Solution:

Area of rectangle = ab

100 = x. b

b = 100/x

Perimeter= 2(a +b)

P = 2a + 2b

P = 2x + 2× 100/x

P = 2x + 200/x

To prove the least perimeter differentiate the perimeter P w.r.t. x,

dp/dx = 2 - 200/x²

Now equate the above function with zero,

2-200/x² = 0

200/x² = 2

x² = 100

x = ± 10

x = -10 is not valid as length can not be negative.

substitute x = 10, in parent function

P = 2x + 200/x

P = 2×10 + 200/10 = 20 + 20 = 40

Hence proved

P (Least perimeter) = 40