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.
⇒ [tex]P = 2(\frac{72}{q} + q)[/tex] {From equation (1)}
Now, condition for perimeter to be maximum is
[tex]\frac{dP}{dq} = 0 = 2(-\frac{72}{q^{2}} + 1 )[/tex]
⇒ q² = 72
⇒ q = 6√2
Therefore, from equation (1) [tex]p = \frac{72}{6\sqrt{2}} = 6\sqrt{2}[/tex].
So, the perimeter, P = 2(6√2 + 6√2) = 24√2. {From equation (2)}
