Question

if the area of a park is 24 sq m then find the dimension for the park which will have the least perimeter ​

Answer 1

hi!

Since, the word perimeter is used instead of circumference; so, I've considered the park to be non-circular; and a quadrilateral.

Let, length of the park = l metres.

So, breadth of the park = (24 / l) metres.

Perimeter of the park = P = 2 * {l + (24 / l)} metres.

Now, for optimum values (maximum or minimum) of P; dP / dl = 0.

Here, dP / dl = [2 * 1 + {48 * (-1) / (l^2)}] = 2 - {48 / (l^2)}.

When dP / dl = 0; 2 * (l^2) = 48; or, l = √24 = 2√6 ≈ 4.89898.

Now, d^2 P / dl^2 = 0 - {48 * (-2) / (l^3)} = {96 / (l^3)} > 0.

So, l ≈ 4.89898 stands for the minimum value of P.

So, for minimum perimeter b = (24 / l) = √24 ≈ 4.89898.

Therefore, perimeter of the park will be minimum when it will be square in shape having dimensions of (√24 metres * √24 metres) ≈ (4.89898 metres * 4.89898 metres).

hope it's help you ✌️..

Answer 2

Answer:

Since, the word perimeter is used instead of circumference; so, I've considered the park to be non-circular; and a quadrilateral.

Let, length of the park = l metres.

So, breadth of the park = (24 / l) metres.

Perimeter of the park = P = 2 * {l + (24 / l)} metres.

Now, for optimum values (maximum or minimum) of P; dP / dl = 0.

Here, dP / dl = [2 * 1 + {48 * (-1) / (l^2)}] = 2 - {48 / (l^2)}.

When dP / dl = 0; 2 * (l^2) = 48; or, l = √24 = 2√6 ≈ 4.89898.

Now, d^2 P / dl^2 = 0 - {48 * (-2) / (l^3)} = {96 / (l^3)} > 0.

So, l ≈ 4.89898 stands for the minimum value of P.

So, for minimum perimeter b = (24 / l) = √24 ≈ 4.89898.

Therefore, perimeter of the park will be minimum when it will be square in shape having dimensions of (√24 metres * √24 metres) ≈ (4.89898 metres * 4.89898 metres).