Question

The area of a rectangle is 81cm^2 what are the dimensions if the perimeter is minimum. What is the value of these minimum perimeter?

Answer 1

Answer:

The perimeter is 36 cm

Step-by-step explanation:

Given;

area of the rectangle, A = 81 cm²

let the Length, = L

let the breadth = b

Perimeter, P = 2L + 2b

Area, A = Lb = 81

$L = \frac{81}{b} \\\\P = 2(\frac{81}{b} ) \ + \ 2b\\\\P = \frac{162}{b} \ + \ 2b\\\\P = 162(b^{-1}) \ + \ 2b\\\\minimize \ the \ perimeter \ to \ obtain \ the \ critical \ points;\\\\P' = -162 (b^{-2}) \ + \ 2\\\\P' = \frac{-162}{b^2} \ + \ 2\\\\P' = \frac{-162 \ + \ 2b^2}{b^2} \\\\at \ critical \ points; P' = 0\\\\\frac{-162 \ + \ 2b^2}{b^2} = 0\\\\-162 \ + \ 2b^2 = 0\\\\2b^2 = 162\\\\b^2 = 81\\\\b = +/- \ \sqrt{81} \\\\b = +/- \ \ 9$

$since \ the \ breadth \ will \ be \ positive \ ; b = 9 \ cm$

L = 81/9

L = 9 cm

The perimeter = 2 (L + b)

                         = 2 (9 + 9)

                         = 2(18)

                         = 36 cm