Question

Hello everyone!

Answer 1

Answer:

Dimensions are 6√2 by 6√2 and perimeter = 24√2.

Step-by-step explanation:

Let p by q gives the dimensions of the rectangle.

So, length = p and width is q.

Now, given that area of the rectangle is, A = pq = 72 ......... (1)

Now, P = 2(p + q) .......... (2) gives the perimeter of the rectangle.  

⇒ $P = 2(\frac{72}{q} + q)$ {From equation (1)}

Now, condition for perimeter to be maximum is  

$\frac{dP}{dq} = 0 = 2(-\frac{72}{q^{2}} + 1 )$

⇒ q² = 72

⇒ q = 6√2

Therefore, from equation (1) $p = \frac{72}{6\sqrt{2}} = 6\sqrt{2}$.

So, the perimeter, P = 2(6√2 + 6√2) = 24√2. {From equation (2)}