Respuesta :
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:
[tex]x\cdot y=16...(1)[/tex]
We know that perimeter of rectangle is sum of its all sides that is:
[tex]P=2x+2y...(2)[/tex]
From equation (1), we will get:
[tex]y=\frac{16}{x}[/tex]
Upon substituting this value in equation (2), we will get:
[tex]P=2x+2(\frac{16}{x})[/tex]
[tex]P=2x+\frac{32}{x}[/tex]
[tex]P=2x+32x^{-1}[/tex]
Now, we will find the derivative of perimeter equation as:
[tex]P'=\frac{d}{dx}(2x)+\frac{d}{dx}(32x^{-1})[/tex]
[tex]P'=2-32x^{-2}[/tex]
Now, we will equate our derivative equal to 0 to find critical points as:
[tex]2-32x^{-2}=0[/tex]
[tex]2-\frac{32}{x^2}=0[/tex]
[tex]-\frac{32}{x^2}=-2[/tex]
[tex]\frac{32}{x^2}=2[/tex]
[tex]2x^2=32[/tex]
[tex]x^2=16[/tex]
Take square root of both sides:
[tex]x=\pm 4[/tex]
Since length cannot be negative, therefore, [tex]x=4[/tex].
Now, we will find 2nd derivative as:
[tex]P''=\frac{d}{dx}(2)-\frac{d}{dx}(32x^{-2})[/tex]
[tex]P''=0-(-2*32x^{-3})[/tex]
[tex]P''=64x^{-3}[/tex]
We know that where 2nd derivative is positive, the point is a minimum. Let us substitute [tex]x=4[/tex] in 2nd derivative.
[tex]P''(4)=64(4)^{-3}[/tex]
[tex]P''(4)=64*\frac{1}{4^3}[/tex]
[tex]P''(4)=64*\frac{1}{64}[/tex]
[tex]P''(4)=1[/tex]
Since [tex]P''(4)=1[/tex], therefore [tex]x=4[/tex] is a minimum.
Let us solve for y using equation [tex]y=\frac{16}{x}[/tex] as:
[tex]y=\frac{16}{4}[/tex]
[tex]y=4[/tex]
Therefore, width and length of 4 inches each will result in smallest perimeter.
[tex]P=2x+2y[/tex]
[tex]P=2(4)+2(4)[/tex]
[tex]P=8+8[/tex]
[tex]P=16[/tex]
Therefore, the smallest possible perimeter would be 16 inches.
