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an open-top box is to be constructed from a sheet of tin that measures 22 inches by 14 inches by cutting out squares from each corner. let V(x) denote the volume of the resulting box. step 1 of 2: write V(x) as a product of linear factorsstep 2 of 2: among the values of x for which V(x)=0, which are physically possible?

Sagot :

It is given that an open-top box is to be constructed from a sheet of tin that measures 22 inches by 14 inches by cutting out squares from each corner

Let x be the measure of the side of the square.

Length of the resulting box =22-2x

Width of the resulting box=14-2x

Height of the resulting box=x

The volume of the box is

[tex]V=\text{length }\times width\times height[/tex]

Substitute values, we get

[tex]V(x)=(22-2x)(14-2x)x[/tex]

[tex]=(22-2x)(14x-2x^2)[/tex]

[tex]=22\mleft(14x-2x^2\mright)-2x\mleft(14x-2x^2\mright)[/tex]

[tex]=22\times14x-22\times2x^2-2x\times14x-(-2x)2x^2[/tex]

[tex]=308x-44x^2-28x^2+4x^3[/tex]

[tex]V(x)=4x^3-72x^2+308x[/tex]

Putting V(x)=0, we get

[tex]4x^3-72x^2+308x=0[/tex]

[tex]4x(x^2-18x+77)=0[/tex]

[tex]4x=0,(x^2-18x+77)=0[/tex]

Here x is not zero

[tex]x^2-18x+77=0[/tex]

[tex]x^2-11x-7x+77=0[/tex]

[tex]x(x^{}-11)-7(x-11)=0[/tex]

[tex](x^{}-11)(x-7)=0[/tex]

[tex](x^{}-11)=0\text{ or }\mleft(x-7\mright)=0[/tex]

[tex]x^{}=11\text{ or }x=7[/tex]

The height of the box is 11 or 7

If the height is 11 inches, substitute x=11 in the length equation, we get

[tex]\text{length =22=2x=22-2}\times11=22-22=0[/tex]

we get a length is 0, so it is not possible to make the box.

Setting x=7, the height of the box is 7 inches.

[tex]\text{Length =22-2x=22-2}\times7=22-14=8inches[/tex]

[tex]\text{width =14-2}\times7=14-14=0[/tex]

we get a width is 0, so it is not possible to make the box.

Hence among the values of x for which V(x)=0 is not physically possible.

View image BerkliS390829
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