An open box is to be formed from a square sheet of metal 3 feet on a side by cutting equal squares out of the corners and bending up the flaps....

Friday, May 4, 2012

An open box is to be formed from a square sheet of metal 3 feet on a side by cutting equal squares out of the corners and bending up the flaps....

You need to create the volume function, hence, you need to write the formula of volume of the box, such that:

V = width x length x height

width = length = $\displaystyle{3}-{2}{x}$

height = x

$\displaystyle{V}={\left({3}-{2}{x}\right)}{\left({3}-{2}{x}\right)}\cdot{x}\Rightarrow{V}={{\left({3}-{2}{x}\right)}}^{{2}}\cdot{x}\Rightarrow{V}={9}{x}-{12}{{x}}^{{2}}+{4}{{x}}^{{3}}$

You need to find the dimensions of the box if the volume is the largest possible, hence, you need to evaluate the derivative of the volume function, such that:

$\displaystyle{V}'{\left({x}\right)}={9}-{24}{x}+{12}{{x}}^{{2}}$

You need to set the derivative function equal to 0, such that:

$\displaystyle{9}-{24}{x}+{12}{{x}}^{{2}}={0}\Rightarrow{3}-{8}{x}+{4}{{x}}^{{2}}={0}$

$\displaystyle{4}{{x}}^{{2}}-{8}{x}+{3}={0}$

Using quadratic formula yields:

$\displaystyle{x}_{{{1},{2}}}=\frac{{{8}\pm\sqrt{{{64}-{48}}}}}{{8}}\Rightarrow{x}_{{{1},{2}}}=\frac{{{8}\pm\sqrt{{16}}}}{{8}}$

$\displaystyle{x}_{{{1},{2}}}=\frac{{{8}\pm{4}}}{{8}}\Rightarrow{x}_{{1}}=\frac{{12}}{{8}}=\frac{{3}}{{2}}={1.5}$

$\displaystyle{x}_{{2}}=\frac{{4}}{{8}}\Rightarrow{x}_{{2}}={0.5}$

You need to select the solution x = 0.5 because x = 1.5 falls outside the domain for x.

width = length = $\displaystyle{3}-{2}{x}={3}-{2}\cdot{0.5}$ => width = length = 1

Hence, evaluating the dimension of the box, under the given conditions, yields width = length = 1 and height = 0.5 ft.

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Friday, May 4, 2012