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?

Expert's answer

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



LATEST TUTORIALS
APPROVED BY CLIENTS