4.8 Lagrange multipliers

Let’s follow the problem-solving strategy:

1. The optimization function is $f\left(x,y\right)=x^2+4y^2-2x+8y.$ To determine the constraint function, we must first subtract $7$ from both sides of the constraint. This gives $x+2y-7=0.$ The constraint function is equal to the left-hand side, so $g\left(x,y\right)=x+2y-7.$ The problem asks us to solve for the minimum value of $f,$ subject to the constraint (see the following graph).
   

   

2. We then must calculate the gradients of both f and g :
   
   $\begin{array}{c}\nabla f\left(x,y\right)=\left(2x-2\right)i+\left(8y+8\right)j\hfill \\ \nabla g\left(x,y\right)=i+2j.\hfill \end{array}$
   
   The equation $\nabla f\left(x_0,y_0\right)=\lambda \nabla g\left(x_0,y_0\right)$ becomes
   
   $\left(2x_0-2\right)i+\left(8y_0+8\right)j=\lambda \left(i+2j\right),$
   
   which can be rewritten as
   
   $\left(2x_0-2\right)i+\left(8y_0+8\right)j=\lambda i+\lambda j.$
   
   Next, we set the coefficients of $i\text{and}j$ equal to each other:
   
   $\begin{array}{c}2x_0-2=\lambda \hfill \\ 8y_0+8=2\lambda .\hfill \end{array}$
   
   The equation $g\left(x_0,y_0\right)=0$ becomes $x_0+2y_0-7=0.$ Therefore, the system of equations that needs to be solved is
   
   $\begin{array}{ccc}\hfill 2x_0-2& =\hfill & \lambda \hfill \\ \hfill 8y_0+8& =\hfill & 2\lambda \hfill \\ \hfill x_0+2y_0-7& =\hfill & 0.\hfill \end{array}$
3. This is a linear system of three equations in three variables. We start by solving the second equation for $\lambda$ and substituting it into the first equation. This gives $\lambda =4y_0+4,$ so substituting this into the first equation gives
   
   $2x_0-2=4y_0+4.$
   
   Solving this equation for $x_0$ gives $x_0=2y_0+3.$ We then substitute this into the third equation:
   $\begin{array}{ccc}\hfill \left(2y_0+3\right)+2y_0-7& =\hfill & 0\hfill \\ \hfill 4y_0-4& =\hfill & 0\hfill \\ \hfill y_0& =\hfill & 1.\hfill \end{array}$
   
   Since $x_0=2y_0+3,$ this gives $x_0=5.$
4. Next, we substitute $\left(5,1\right)$ into $f\left(x,y\right)=x^2+4y^2-2x+8y,$ gives $f\left(5,1\right)=5^2+4{\left(1\right)}^2-2\left(5\right)+8\left(1\right)=27.$ To ensure this corresponds to a minimum value on the constraint function, let’s try some other values, such as the intercepts of $g\left(x,y\right)=0,$ Which are $\left(7,0\right)$ and $\left(0,3.5\right).$ We get $f\left(7,0\right)=35$ and $f\left(0,3.5\right)=77,$ so it appears $f$ has a minimum at $\left(5,1\right).$