Question

Find the dimensions of a rectangle (in m) with area 1,331 m2 whose perimeter is as small as possible. (Enter the dimensions as a comma separated list.)

Answer 1

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;

$F = 2x + 2y\\\\F = 2x + 2(\frac{1331}{x} )\\\\F = 2x + \frac{2662}{x}$

Determine the derivative of the function;

$F' = 2 - \frac{2662}{x^2}$

Find the critical points;

F' = 0

$2 - \frac{2662}{x^2} = 0\\\\\frac{2662}{x^2} = 2\\\\x^2 = \frac{2662}{2} \\\\x^2 = 1331\\\\x =\sqrt{1331}\\\\ x = 36.48 \ m$

Take the second derivative of the function

$F'' = \frac{5324}{x^3}\\\\ \frac{5324}{x^3} >0\\\\thus, the \ perimeter \ will \ be \ minimum \ at \ x = 36.48$

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

Answer 2

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:

$\mathbf{A = 1331}$

Let the dimension be x and y.

So, we have:

$\mathbf{A = xy = 1331}$

Make x the subject

$\mathbf{x = \frac{1331}{y}}$

The perimeter is calculated as:

$\mathbf{P = 2(x + y)}$

Substitute $\mathbf{x = \frac{1331}{y}}$

$\mathbf{P = 2(\frac{1331}{y} + y)}$

Expand

$\mathbf{P = \frac{2662}{y} + 2y}$

Differentiate

$\mathbf{P' = -\frac{2662}{y^2} + 2}$

Set to 0

$\mathbf{ -\frac{2662}{y^2} + 2 = 0}$

Rewrite as:

$\mathbf{ -\frac{2662}{y^2} = -2}$

Divide both sides by -1

$\mathbf{ \frac{2662}{y^2} = 2}$

Multiply y^2

$\mathbf{2662 = 2y^2}$

Divide by 2

$\mathbf{1331 = y^2}$

Take square roots of both sides

$\mathbf{y = \sqrt{1331}}$

$\mathbf{y = 36.5}$

Substitute $\mathbf{y = \sqrt{1331}}$ in $\mathbf{x = \frac{1331}{y}}$

$\mathbf{x = \frac{1331}{\sqrt{1331}}}$

$\mathbf{x = \sqrt{1331}}$

$\mathbf{x = 36.5}$

Hence, the dimensions that minimize the perimeter are 36.5, 36.5

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