find the smallest perimeter and the dimensions for a rectangle with an area of 4 in²\n\nthe smallest…

Explanation:

Step1: Let the length be (x) and width be (y)

We know that the area (A = xy=4), so (y=\frac{4}{x}). The perimeter (P = 2(x + y)=2\left(x+\frac{4}{x}\right)), (x>0)

Step2: Find the derivative of (P(x))

Differentiate (P(x)=2x+\frac{8}{x}) with respect to (x). Using the power rule ((x^n)^\prime=nx^{n - 1}), (P^\prime(x)=2-\frac{8}{x^{2}})

Step3: Find the critical points

Set (P^\prime(x) = 0), then (2-\frac{8}{x^{2}}=0). Cross - multiply: (2x^{2}-8 = 0), (x^{2}=4), (x = 2) (since (x>0))

Step4: Check the second - derivative

Differentiate (P^\prime(x)) to get (P^{\prime\prime}(x)=\frac{16}{x^{3}}). When (x = 2), (P^{\prime\prime}(2)=\frac{16}{8}=2>0), so (P(x)) has a minimum at (x = 2)

Step5: Find the dimensions

When (x = 2), (y=\frac{4}{2}=2)

Answer:

(2,2)