Find the dimensions of a rectangle (in m) with area 1,000 m2 whose perimeter is as small as possible. (Enter the dimensions as a

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3 years ago

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1 answer:

Amanda [17] · Answer 1 · 2022-08-06T15:12:58+03:00

The perimeter of the rectangle is the sum of its dimensions

The dimensions that minimize the perimeter are $\mathbf{10\sqrt{10 },10\sqrt{10 }}$

The area is given as:

$\mathbf{A = 1000}$

Let the dimension be x and y.

So, we have:

$\mathbf{A = xy = 1000}$

Make x the subject

$\mathbf{x = \frac{1000}{y}}$

The perimeter is calculated as:

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

Substitute $\mathbf{x = \frac{1000}{y}}$

$\mathbf{P = 2(\frac{1000}{y} + y)}$

Expand

$\mathbf{P = \frac{2000}{y} + 2y}$

Differentiate

$\mathbf{P' = -\frac{2000}{y^2} + 2}$

Set to 0

$\mathbf{ -\frac{2000}{y^2} + 2 = 0}$

Rewrite as:

$\mathbf{ -\frac{2000}{y^2} = -2}$

Divide both sides by -1

$\mathbf{\frac{2000}{y^2} = 2}$

Multiply y^2

$\mathbf{2000 = 2y^2}$

Divide by 2

$\mathbf{1000 = y^2}$

Take square roots of both sides

$\mathbf{y = \sqrt{1000 }}$

$\mathbf{y = 10\sqrt{10 }}$

Substitute $\mathbf{y = \sqrt{1000 }}$ in $\mathbf{x = \frac{1000}{y}}$

$\mathbf{x = \frac{1000}{\sqrt{1000}}}$

$\mathbf{x = \sqrt{1000}}$

$\mathbf{x = 10\sqrt{10 }}$

Hence, the dimensions that minimize the perimeter are $\mathbf{10\sqrt{10 },10\sqrt{10 }}$