Question

What is the smallest perimeter possible for a rectangle whose area is 16 in^2, and what are its dimensions?

Answer 1

Answer:

Smallest possible perimeter 16 inches.

Length : 4 inches,

Width : 4 inches.

Step-by-step explanation:

Let x represent length and y represent width of rectangle.

We have been given that area of a rectangle is 16 square inches.

We know that area of rectangle is product of its length and width. We can represent our given information in an equation as:

$x\cdot y=16...(1)$

We know that perimeter of rectangle is sum of its all sides that is:

$P=2x+2y...(2)$

From equation (1), we will get:

$y=\frac{16}{x}$

Upon substituting this value in equation (2), we will get:

$P=2x+2(\frac{16}{x})$

$P=2x+\frac{32}{x}$

$P=2x+32x^{-1}$

Now, we will find the derivative of perimeter equation as:

$P'=\frac{d}{dx}(2x)+\frac{d}{dx}(32x^{-1})$

$P'=2-32x^{-2}$  

Now, we will equate our derivative equal to 0 to find critical points as:

$2-32x^{-2}=0$

$2-\frac{32}{x^2}=0$

$-\frac{32}{x^2}=-2$

$\frac{32}{x^2}=2$

$2x^2=32$

$x^2=16$

Take square root of both sides:

$x=\pm 4$

Since length cannot be negative, therefore, $x=4$.

Now, we will find 2nd derivative as:

$P''=\frac{d}{dx}(2)-\frac{d}{dx}(32x^{-2})$  

$P''=0-(-2*32x^{-3})$  

$P''=64x^{-3}$  

We know that where 2nd derivative is positive, the point is a minimum. Let us substitute $x=4$ in 2nd derivative.

$P''(4)=64(4)^{-3}$

$P''(4)=64*\frac{1}{4^3}$

$P''(4)=64*\frac{1}{64}$

$P''(4)=1$

Since $P''(4)=1$, therefore  $x=4$ is a minimum.

Let us solve for y using equation $y=\frac{16}{x}$ as:

$y=\frac{16}{4}$

$y=4$

Therefore, width and length of 4 inches each will result in smallest perimeter.

$P=2x+2y$

$P=2(4)+2(4)$

$P=8+8$

$P=16$

Therefore, the smallest possible perimeter would be 16 inches.