4.8 Lagrange multipliers  (Page 3/11)

Again, we follow the problem-solving strategy:

1. The optimization function is $f\left(x,y\right)=48x+96y-x^2-2xy-9y^2.$ To determine the constraint function, we first subtract 216 from both sides of the constraint, then divide both sides by $4,$ which gives $5x+y-54=0.$ The constraint function is equal to the left-hand side, so $g\left(x,y\right)=5x+y-54.$ The problem asks us to solve for the maximum value of $f,$ subject to this constraint.
2. So, we calculate the gradients of both $f\text{and}$ $g\text{:}$
   
   $\begin{array}{c}\nabla f\left(x,y\right)=\left(48-2x-2y\right)i+\left(96-2x-18y\right)j\hfill \\ \nabla g\left(x,y\right)=5i+j.\hfill \end{array}$
   
   The equation $\nabla f\left(x_0,y_0\right)=\lambda \nabla g\left(x_0,y_0\right)$ becomes
   
   $\left(48-2x_0-2y_0\right)i+\left(96-2x_0-18y_0\right)j=\lambda \left(5i+j\right),$
   
   which can be rewritten as
   
   $\left(48-2x_0-2y_0\right)i+\left(96-2x_0-18y_0\right)j=\lambda 5i+\lambda j.$
   
   We then set the coefficients of $i\text{and}j$ equal to each other:
   
   $\begin{array}{ccc}\hfill 48-2x_0-2y_0& =\hfill & 5\lambda \hfill \\ \hfill 96-2x_0-18y_0& =\hfill & \lambda .\hfill \end{array}$
   
   The equation $g\left(x_0,y_0\right)=0$ becomes $5x_0+y_0-54=0.$ Therefore, the system of equations that needs to be solved is
   
   $\begin{array}{ccc}\hfill 48-2x_0-2y_0& =\hfill & 5\lambda \hfill \\ \hfill 96-2x_0-18y_0& =\hfill & \lambda \hfill \\ \hfill 5x_0+y_0-54& =\hfill & 0.\hfill \end{array}$
3. We use the left-hand side of the second equation to replace $\lambda$ in the first equation:
   
   $\begin{array}{ccc}\hfill 48-2x_0-2y_0& =\hfill & 5\left(96-2x_0-18y_0\right)\hfill \\ \hfill 48-2x_0-2y_0& =\hfill & 480-10x_0-90y_0\hfill \\ \hfill 8x_0& =\hfill & 432-88y_0\hfill \\ \hfill x_0& =\hfill & 54-11y_0.\hfill \end{array}$
   
   Then we substitute this into the third equation:
   
   $\begin{array}{ccc}\hfill 5\left(54-11y_0\right)+y_0-54& =\hfill & 0\hfill \\ \hfill 270-55y_0+y_0& =\hfill & 0\hfill \\ \hfill 216-54y_0& =\hfill & 0\hfill \\ \hfill y_0& =\hfill & 4.\hfill \end{array}$
   
   Since $x_0=54-11y_0,$ this gives $x_0=10.$
4. We then substitute $\left(10,4\right)$ into $f\left(x,y\right)=48x+96y-x^2-2xy-9y^2,$ which gives
   
   $\begin{array}{cc}\hfill f\left(5,1\right)& =48\left(10\right)+96\left(4\right)-{\left(10\right)}^2-2\left(10\right)\left(4\right)-9{\left(4\right)}^2\hfill \\ & =480+384-100-80-144=540.\hfill \end{array}$
   
   Therefore the maximum profit that can be attained, subject to budgetary constraints, is $\text{\$}\mathrm{540,000}$ with a production level of $\mathrm{10,000}$ golf balls and $4$ hours of advertising bought per month. Let’s check to make sure this truly is a maximum. The endpoints of the line that defines the constraint are $\left(10.8,0\right)$ and $\left(0,54\right)$ Let’s evaluate $f$ at both of these points:
   
   $\begin{array}{ccc}\hfill f\left(10.8,0\right)& =\hfill & 48\left(10.8\right)+96\left(0\right)-{10.8}^2-2\left(10.8\right)\left(0\right)-9\left(0^2\right)=401.76\hfill \\ \hfill f\left(0,54\right)& =\hfill & 48\left(0\right)+96\left(54\right)-0^2-2\left(0\right)\left(54\right)-9\left({54}^2\right)=\mathrm{-21},060.\hfill \end{array}$
   
   The second value represents a loss, since no golf balls are produced. Neither of these values exceed $540,$ so it seems that our extremum is a maximum value of $f.$