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Nady [450]
3 years ago
15

A bin is constructed from sheet metal with a square base and 4 equal rectangular sides. if the bin is constructed from 48 square

feet of sheet metal, then what is the largest volume of such a bin? what are its dimensions?

Mathematics
1 answer:
kondaur [170]3 years ago
8 0
This is a problem of maxima and minima using derivative.

In the figure shown below we have the representation of this problem, so we know that the base of this bin is square. We also know that there are four square rectangles sides. This bin is a cube, therefore the volume is:

V = length x width x height

That is:

V = xxy = x^{2}y

We also know that the <span>bin is constructed from 48 square feet of sheet metal, s</span>o:

Surface area of the square base = x^{2}

Surface area of the rectangular sides = 4xy

Therefore, the total area of the cube is:

A = 48 ft^{2} =  x^{2} + 4xy

Isolating the variable y in terms of x:

y =  \frac{48- x^{2} }{4x}

Substituting this value in V:

V =  x^{2}( \frac{48- x^{2} }{x}) = 48x- x^{3}

Getting the derivative and finding the maxima. This happens when the derivative is equal to zero:

\frac{dv}{dx} = 48-3x^{2} =0

Solving for x:

x =  \sqrt{\frac{48}{3}} =  \sqrt{16} = 4

Solving for y:

y =  \frac{48- 4^{2} }{(4)(4)} = 2

Then, <span>the dimensions of the largest volume of such a bin is:
</span>
Length = 4 ft
Width =  4 ft
Height = 2 ft

And its volume is:

V = (4^{2} )(2) = 32 ft^{3}

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34 + 4x -2 = 180

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Given right triangle ABCABC with altitude \overline{BD} BD drawn to hypotenuse ACAC. If BC=10BC=10 and DC=5,DC=5, what is the le
Helen [10]

Answer: \overline{AC}= 12.5

Step-by-step explanation: The triangle ABC and its altitude \overline{BD} is represented in the figure below.

<u>Altitude</u> is a segment of line that link a vertex and the opposite side, forming a right angle.

So, because of \overline{BD}, now we have two similar triangles, which means that ratios of corresponding sides are equal:

\frac{\overline{BD}}{\overline{BC}} =\frac{\overline{AD}}{\overline{BD}}

\overline{BD}^{2}=\overline{BC}.\overline{AD} (1)

This is always true for a right triangle and a altitude drawn to the hypotenuse.

Triangle BDC is also right triangle. So, we can use Pythagorean theorem to determine the missing side.

\overline{BC}^{2}=\overline{CD}^{2}+\overline{BD}^{2}

\overline{BD}^{2}=\overline{BC}^{2}-\overline{CD}^{2} (2)

Substituting (2) into (1):

\overline{BC}^{2}-\overline{CD}^{2}=\overline{BC}.\overline{AD}

We want to find f, so:

\overline{AD}=\frac{\overline{BC}^{2}-\overline{CD}^{2}}{\overline{BC}}

f=\frac{10^{2}-5^{2}}{10}

f = 7.5

The length of \overline{AC} is

\overline{AC}=f+5

\overline{AC}=7.5+5

\overline{AC}=12.5

The length of the hypotenuse of triangle ABC is 12.5 units.

5 0
2 years ago
Read 2 more answers
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