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vagabundo [1.1K]
3 years ago
7

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

comma separated list.)
Mathematics
1 answer:
Amanda [17]3 years ago
8 0

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 }}

Read more about perimeters at:

brainly.com/question/6465134

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Step-by-step explanation:

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Since we only have even numbers to work with, the ones digit must be an even multiple of 2: 4 or 8. (The tens digit cannot be odd.) The digits 4 and 8 comprise half of the available digits, so half of all possible numbers made from these digits will be divisible by 4.

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