Revenue function

\(\displaystyle{R}{\left({x},{y}\right)}={18}{x}+{25}{y}\)

Profit function

\(\displaystyle{C}{\left({x},{y}\right)}={R}{\left({x},{y}\right)}-{C}{\left({x},{y}\right)}\)

\(\displaystyle{C}{\left({x},{y}\right)}=\cdot{18}{x}+{25}{y}{)}-{\left({4}{x}^{{2}}-{6}{x}{y}+{3}{y}^{{2}}+{20}{x}+{19}{y}-{12}\right)}\)

\(\displaystyle{C}{\left({x},{y}\right)}=-{4}{x}^{{2}}+{6}\times{y}-{3}{y}^{{2}}-{2}{x}+{6}{y}+{12}\)

First we will find the critical point. Now,

\(\displaystyle{P}_{{x}}{\left({x},{y}\right)}=-{8}{x}+{6}{y}-{2}\)

\(\displaystyle{P}_{{y}}{\left({x},{y}\right)}={6}{x}-{6}{y}+{6}\)

\(\displaystyle{P}_{{\times}}{\left({x},{y}\right)}=-{8}\)

\(\displaystyle{P}_{{{y}{y}}}{\left({x},{y}\right)}=-{6}\)

\(\displaystyle{P}_{{y}}{\left({x},{y}\right)}={0}\)

6x-6y+6=0

y=x+1

And \(\displaystyle{P}_{{x}}{\left({x},{y}\right)}={0}\)

6y-8x-2=0

6(x+1)-8x=2

x=2 (i)

So, y=x+1

y=2 (ii)

Hence, the critical point is (2,3)

Since \(\displaystyle{f}_{{\times}},{f}_{{{y}{y}}},{\quad\text{and}\quad}{f}_{{{x}{y}}}\) all are constant so we will need to pkug critical point to function

\(\displaystyle{D}{\left({x},{y}\right)}={f}_{{\times}}{f}_{{{y}{y}}}-{{f}_{{{x}{y}}}^{{2}}}\)

\(\displaystyle={\left(-{8}\right)}{\left(-{6}\right)}-{\left(-{6}\right)}^{{2}}\)

=12>0

Since, D>0, and \(\displaystyle{f}_{{\times}}{<}{0}\)</span>

The critical point P(2,3) has a relative maximum.

And the maximum value of profit function

\(\displaystyle{P}{\left({2},{3}\right)}=-{4}{\left({2}\right)}^{{2}}+{6}\cdot{2}\cdot{3}-{3}{\left({3}\right)}^{{2}}-{2}\cdot{2}+{6}\cdot{3}+{12}\)

\(\displaystyle{P}{\left({2},{3}\right)}={19}\)

The maximum profit of $19000 will be earned if 2000 shirt of $18 and 3000 shirts of $25 are produced and sold.

\(\displaystyle{R}{\left({x},{y}\right)}={18}{x}+{25}{y}\)

Profit function

\(\displaystyle{C}{\left({x},{y}\right)}={R}{\left({x},{y}\right)}-{C}{\left({x},{y}\right)}\)

\(\displaystyle{C}{\left({x},{y}\right)}=\cdot{18}{x}+{25}{y}{)}-{\left({4}{x}^{{2}}-{6}{x}{y}+{3}{y}^{{2}}+{20}{x}+{19}{y}-{12}\right)}\)

\(\displaystyle{C}{\left({x},{y}\right)}=-{4}{x}^{{2}}+{6}\times{y}-{3}{y}^{{2}}-{2}{x}+{6}{y}+{12}\)

First we will find the critical point. Now,

\(\displaystyle{P}_{{x}}{\left({x},{y}\right)}=-{8}{x}+{6}{y}-{2}\)

\(\displaystyle{P}_{{y}}{\left({x},{y}\right)}={6}{x}-{6}{y}+{6}\)

\(\displaystyle{P}_{{\times}}{\left({x},{y}\right)}=-{8}\)

\(\displaystyle{P}_{{{y}{y}}}{\left({x},{y}\right)}=-{6}\)

\(\displaystyle{P}_{{y}}{\left({x},{y}\right)}={0}\)

6x-6y+6=0

y=x+1

And \(\displaystyle{P}_{{x}}{\left({x},{y}\right)}={0}\)

6y-8x-2=0

6(x+1)-8x=2

x=2 (i)

So, y=x+1

y=2 (ii)

Hence, the critical point is (2,3)

Since \(\displaystyle{f}_{{\times}},{f}_{{{y}{y}}},{\quad\text{and}\quad}{f}_{{{x}{y}}}\) all are constant so we will need to pkug critical point to function

\(\displaystyle{D}{\left({x},{y}\right)}={f}_{{\times}}{f}_{{{y}{y}}}-{{f}_{{{x}{y}}}^{{2}}}\)

\(\displaystyle={\left(-{8}\right)}{\left(-{6}\right)}-{\left(-{6}\right)}^{{2}}\)

=12>0

Since, D>0, and \(\displaystyle{f}_{{\times}}{<}{0}\)</span>

The critical point P(2,3) has a relative maximum.

And the maximum value of profit function

\(\displaystyle{P}{\left({2},{3}\right)}=-{4}{\left({2}\right)}^{{2}}+{6}\cdot{2}\cdot{3}-{3}{\left({3}\right)}^{{2}}-{2}\cdot{2}+{6}\cdot{3}+{12}\)

\(\displaystyle{P}{\left({2},{3}\right)}={19}\)

The maximum profit of $19000 will be earned if 2000 shirt of $18 and 3000 shirts of $25 are produced and sold.