Answer:
The dimensions of the rectangle are x = 36.48 m and y = 36.48 m
Step-by-step explanation:
Let the sides of the rectangle be x and y
Objective function is given by;
F = 2(x + y)
F = 2x + 2y
The area of the rectangle is given by;
xy = 1331
y = 1331 / x
Substitute in the value of y in objective function;
[tex]F = 2x + 2y\\\\F = 2x + 2(\frac{1331}{x} )\\\\F = 2x + \frac{2662}{x}\\\\[/tex]
Determine the derivative of the function;
[tex]F' = 2 - \frac{2662}{x^2}[/tex]
Find the critical points;
F' = 0
[tex]2 - \frac{2662}{x^2} = 0\\\\\frac{2662}{x^2} = 2\\\\x^2 = \frac{2662}{2} \\\\x^2 = 1331\\\\x =\sqrt{1331}\\\\ x = 36.48 \ m[/tex]
Take the second derivative of the function
[tex]F'' = \frac{5324}{x^3}\\\\ \frac{5324}{x^3} >0\\\\thus, the \ perimeter \ will \ be \ minimum \ at \ x = 36.48[/tex]
Determine the second dimension;
y = 1331 / x
y = 1331 / 36.48
y = 36.48m
Therefore, the dimensions of the rectangle is x = y = 36.48 m
The perimeter of the rectangle is the sum of its dimensions
The dimensions that minimize the perimeter are 36.5, 36.5
The area is given as:
[tex]\mathbf{A = 1331}[/tex]
Let the dimension be x and y.
So, we have:
[tex]\mathbf{A = xy = 1331}[/tex]
Make x the subject
[tex]\mathbf{x = \frac{1331}{y}}[/tex]
The perimeter is calculated as:
[tex]\mathbf{P = 2(x + y)}[/tex]
Substitute [tex]\mathbf{x = \frac{1331}{y}}[/tex]
[tex]\mathbf{P = 2(\frac{1331}{y} + y)}[/tex]
Expand
[tex]\mathbf{P = \frac{2662}{y} + 2y}[/tex]
Differentiate
[tex]\mathbf{P' = -\frac{2662}{y^2} + 2}[/tex]
Set to 0
[tex]\mathbf{ -\frac{2662}{y^2} + 2 = 0}[/tex]
Rewrite as:
[tex]\mathbf{ -\frac{2662}{y^2} = -2}[/tex]
Divide both sides by -1
[tex]\mathbf{ \frac{2662}{y^2} = 2}[/tex]
Multiply y^2
[tex]\mathbf{2662 = 2y^2}[/tex]
Divide by 2
[tex]\mathbf{1331 = y^2}[/tex]
Take square roots of both sides
[tex]\mathbf{y = \sqrt{1331}}[/tex]
[tex]\mathbf{y = 36.5}[/tex]
Substitute [tex]\mathbf{y = \sqrt{1331}}[/tex] in [tex]\mathbf{x = \frac{1331}{y}}[/tex]
[tex]\mathbf{x = \frac{1331}{\sqrt{1331}}}[/tex]
[tex]\mathbf{x = \sqrt{1331}}[/tex]
[tex]\mathbf{x = 36.5}[/tex]
Hence, the dimensions that minimize the perimeter are 36.5, 36.5
Read more about perimeters at:
https://brainly.com/question/6465134
