find the smallest perimeter and the dimensions for a rectangle with an area of 4 in.²\nthe smallest…

Explanation:

Step1: Define variables

Let the length of the rectangle be (x) inches and the width be (y) inches. Given that the area (A = xy=4), so (y=\frac{4}{x}). The perimeter (P) of a rectangle is (P = 2(x + y)). Substitute (y=\frac{4}{x}) into the perimeter formula, we get (P(x)=2\left(x+\frac{4}{x}\right)), where (x>0).

Step2: Find the derivative of the perimeter function

Using the sum rule ((u + v)'=u'+v') and the power rule ((x^n)'=nx^{n - 1}), if (u = x) and (v=\frac{4}{x}=4x^{-1}). Then (P'(x)=2\left(1-4x^{-2}\right)=2\left(1-\frac{4}{x^{2}}\right)=\frac{2(x^{2}-4)}{x^{2}}).

Step3: Find the critical points

Set (P'(x) = 0), so (\frac{2(x^{2}-4)}{x^{2}}=0). Since (x^{2}>0) for (x>0), we solve (x^{2}-4=0). Factoring, ((x - 2)(x + 2)=0). Since (x>0), (x = 2).

Step4: Use the second - derivative test

Find the second derivative (P''(x)). (P'(x)=2-\frac{8}{x^{2}}), then (P''(x)=\frac{16}{x^{3}}). When (x = 2), (P''(2)=\frac{16}{8}=2>0). So (x = 2) is a point of minimum. When (x = 2), (y=\frac{4}{2}=2).

Step5: Calculate the perimeter

Substitute (x = 2) and (y = 2) into the perimeter formula (P=2(x + y)). (P=2(2 + 2)=8).

Answer:

(8)