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.