# Analysis of daily output of a factory during a 8-hour shift shows that

Analysis of daily output of a factory during a 8-hour shift shows that the hourly number of units y produced after t hours of production is
$$\displaystyle{y}={140}{t}+{\frac{{{1}}}{{{2}}}}{t}^{{{2}}}-{t}^{{{3}}},\ {0}\le{t}\le{8}$$
a) After how many hours will the hourly number of units be maximized? hr
b) What is the maximum hourly output? $$\displaystyle{\frac{{{u}{n}{i}{t}{s}}}{{{h}{r}}}}$$

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Step 1
The number of units proced after t hours of a production is:
$$\displaystyle{y}={140}{t}+{\frac{{{1}}}{{{2}}}}{t}^{{{2}}}-{t}^{{{3}}}$$
Where, $$\displaystyle{0}\le{t}\le{8}$$
Step 2
First order derivative of y with respect to t is:
$$\displaystyle{\frac{{{\left.{d}{y}\right.}}}{{{\left.{d}{t}\right.}}}}={140}{\frac{{{d}}}{{{\left.{d}{t}\right.}}}}{\left({t}\right)}+{\frac{{{1}}}{{{2}}}}{\frac{{{d}}}{{{\left.{d}{t}\right.}}}}{\left({t}^{{{2}}}\right)}-{\frac{{{d}}}{{{\left.{d}{t}\right.}}}}{\left({t}^{{{3}}}\right)}$$
$$\displaystyle={140}{\left({1}\right)}+{\frac{{{1}}}{{{2}}}}{\left({2}{t}\right)}-{3}{t}^{{{2}}}{\left[\because{\frac{{{d}}}{{{\left.{d}{x}\right.}}}}{\left({x}^{{{n}}}\right)}={n}{x}^{{{n}-{1}}}\right]}$$
$$\displaystyle={140}+{t}-{3}{t}^{{{2}}}$$
Second order derivative of y with respect to t is:
$$\displaystyle{\frac{{{d}^{{{2}}}{y}}}{{{\left.{d}{t}\right.}^{{{2}}}}}}={\frac{{{d}}}{{{\left.{d}{t}\right.}}}}{\left({140}\right)}+{\frac{{{d}}}{{{\left.{d}{t}\right.}}}}{\left({t}\right)}-{3}{\frac{{{d}}}{{{\left.{d}{t}\right.}}}}{\left({t}^{{{2}}}\right)}$$
$$\displaystyle={0}+{1}-{3}{\left({2}{t}\right)}$$
$$\displaystyle={1}-{6}{t}$$ To maximazed the units, we have to
$$\displaystyle{s}{e}{t}{\frac{{{\left.{d}{y}\right.}}}{{{\left.{d}{t}\right.}}}}={0}$$
$$\displaystyle\Rightarrow{140}+{t}-{3}{t}^{{{2}}}={0}$$
$$\displaystyle\Rightarrow{3}{t}^{{{2}}}-{t}-{140}={0}$$
$$\displaystyle\Rightarrow{t}={\frac{{-{\left(-{1}\right)}\pm\sqrt{{{\left(-{1}\right)}^{{{2}}}-{4}{\left({3}\right)}{\left(-{140}\right)}}}}}{{{2}{\left({3}\right)}}}}$$
$$\displaystyle\Rightarrow{t}={\frac{{{1}\pm\sqrt{{{1}+{1680}}}}}{{{6}}}}$$
$$\displaystyle\Rightarrow{t}={\frac{{{1}\pm{41}}}{{{6}}}}$$
So, $$\displaystyle{t}={\frac{{{1}+{41}}}{{{6}}}}={7}$$
And, $$\displaystyle{t}={\frac{{{1}-{41}}}{{{6}}}}=-{\frac{{{40}}}{{{6}}}}\approx-{6.67}$$
Step 3
Since t represents time so it can not be negative.
$$\displaystyle\therefore{t}={7}$$
At $$\displaystyle{t}={7}$$
$$\displaystyle{\frac{{{d}^{{{2}}}{y}}}{{{\left.{d}{t}\right.}^{{{2}}}}}}{B}{i}{g}{\mid}_{{{t}={7}}}={1}-{6}{\left({7}\right)}=-{41}{<}{0}$$
Hence, we can say that at $$\displaystyle{t}={7}$$, y is maximized.
a) After 7 hours the hourly number of units be maximized.
b) Now, at $$\displaystyle{t}={7}$$
$$\displaystyle{y}={140}{\left({7}\right)}+{\frac{{{7}^{{{2}}}}}{{{2}}}}-{7}^{{{3}}}$$
$$\displaystyle={980}+{24.5}-{343}$$
$$\displaystyle={661.5}$$
Hence, the maximum hourly output is 661.5 units.