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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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