Question #160108

 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 folding up the sides as shown in the diagram: If 𝑙 = 200𝑚𝑚 and 𝑤 = 150𝑚𝑚 calculate: a) The value of x 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
2021-02-02T05:04:03-0500

Solution

a) The value of x which will give the maximum volume


Given that L:= 200 ; W:= 150


The volume of the box is V=lwhV=lwh, where length l=L2xl=L-2x, width w=W2xw=W-2x, and height h=x. Therefore 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 image below.

Answer:


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


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