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Ira Lisetskai [31]
2 years ago
4

A rectangle is formed by placing two identical squares side by side

Mathematics
2 answers:
prohojiy [21]2 years ago
6 0
Answer: is 162 cm^2.

explanation:

divide the perimeter of the rectangle by 2. 72/2 = 36. 36 cm is the perimeter of one square.

divide 36 by 4. this will give you the side length of the square. 9 cm is the side length of one square.

now use the area formula to find the area of the square. multiply the side length by the side length. 9 x 9 = 81 cm^2.

81 is the area of one square, but since there are two squares, multiply 81 x 2.

you get 162 cm^2.
ANEK [815]2 years ago
4 0

Answer:

288 cm^2

Step-by-step explanation:

If you have 2 congruent squares, side by side, the perimeter is 6*sidelength.

We divide 72 by 6 to find sidelength, and get 12.

We square 12 to find the area of a square, and get 144.

Multiply by 2 to find the area of the rectangle, and get 288 cm^2

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Answer:

l = \sqrt[3]{\frac{2V}{7}}     b = \sqrt[3]{\frac{2V}{7}}       h = \sqrt[3]{\frac{49V}{4}}

Step-by-step explanation:

Represent the volume of the box with V and the dimensions with l, b and h.

The volume (V) is:

V = l * b * h

Make h the subject of the formula

h = \frac{V}{lb}

The surface area (S) of the aquarium is:

S = lb + 2(lh + bh)

Where lb represents the area of the base (i.e. slate):

The cost (C) of the surface area is:

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Differentiate with respect to l and with respect to b

C_l=7b - \frac{2V}{l^2} =0

C_b=7l - \frac{2V}{b^2} =0

To solve for b and l, we equate both equations and set l to b (to minimize the cost)

7b - \frac{2V}{l^2}=7l - \frac{2V}{b^2}

7l - \frac{2V}{l^2}=7b - \frac{2V}{b^2}

By comparison:

l =b

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7l - \frac{2V}{l^2}=0

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Cross Multiply

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Solve for l

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l = \sqrt[3]{\frac{2V}{7}}

Recall that: l =b

b = \sqrt[3]{\frac{2V}{7}}

Also recall that:

h = \frac{V}{lb}

h = \frac{V}{\sqrt[3]{\frac{2V}{7}}*\sqrt[3]{\frac{2V}{7}}}

h = \frac{V}{\sqrt[3]{\frac{4V^2}{49}}}

Apply law of indices

h = \sqrt[3]{\frac{49V^3}{4V^2}}

h = \sqrt[3]{\frac{49V}{4}}

The dimension that minimizes the cost of material of the aquarium is:

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