Question #99173
Question: MUST be answered in analytical format if possible?
You plan to make a simple, open topped box from a piece of sheet metal by cutting a square - of equal size - from each corner and fold up the sides.
If L (length) is 200mm and W (width) is 150mm, calculate:

A) the value of x (x is the value of each corner) which will give the maximum volume?

B) the maximum volume of the box?

C) comment of the value obtained in part b?
1
Expert's answer
2019-11-22T12:06:40-0500

The volume of the box can be written in the form: V(x)=(L2x)(W2x)xV(x)=(L - 2\cdot x)\cdot(W- 2\cdot x)\cdot x

Lengths and width of the box decreased that is of sheet metal by xx from each corner, and height of the box is equal xx. We bring V(x)V(x) to a simple form: V(x)=4x32(L+W)x2+LWxV(x)=4\cdot x^3 -2\cdot (L+W)\cdot x^2+ L\cdot W\cdot x

To find maximum volume one compute the derivative of volume with respect to xx

Vx=12x24(L+W)x+LWV^{'}_x=12\cdot x^2 - 4\cdot (L+W)\cdot x+ L\cdot W and define the root of the equation Vx=0V^{'}_x=0 :

x1,2=(2(L+W)±4(L+W)212LW)/12=16(L+W±L2+W2LW)x_{1,2}=(2\cdot(L+W)\pm\sqrt{4(L+W)^2-12\cdot L\cdot W} )/12=\frac{1}{6}(L+W\pm\sqrt{L^2+W^2-L\cdot W})

x1=88.38; x2=28.29x_1=88.38 ;\space x_2=28.29

The first value cannot be implemented. It is clear that the box will succeed only if x<W/2x<W/2 . The second value corresponds to the maximum volume shown in the figure.

Answer: x=28.29mm;maxV=379037.81mm2x=28.29 mm; max V=379037.81 mm^2



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