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Hello everyone!

can you please help me with this math activity? Help would be very appreciated.


The area of a rectangle is
72^(2). State the dimensions of the rectangle with the least perimeter. Calculate the perimeter.

User Wolfish
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1 Answer

4 votes

Answer:

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

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((72)/(q) + q) {From equation (1)}

Now, condition for perimeter to be maximum is


(dP)/(dq) = 0 = 2(-(72)/(q^(2)) + 1 )

⇒ q² = 72

q = 6√2

Therefore, from equation (1)
p = (72)/(6√(2)) = 6√(2).

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

User Xcorat
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