Answer
Method 1. using Lagrange multipliers
Method 2. setting $y = 5x - 4$ in $f\left( {x,y} \right)$
The minimum value of $f\left( {x,y} \right) = xy$ subject to the constraint $5x - y = 4$ is $ - \frac{4}{5}$.
Work Step by Step
We have $f\left( {x,y} \right) = xy$ subject to the constraint $g\left( {x,y} \right) = 5x - y - 4 = 0$.
Method 1. using Lagrange multipliers
Step 1. Write out the Lagrange equations
Using Theorem 1, the Lagrange condition $\nabla f = \lambda \nabla g$ yields
$\left( {y,x} \right) = \lambda \left( {5, - 1} \right)$
$y = 5\lambda $, ${\ \ \ \ }$ $x = - \lambda $
Step 2. Solve for $\lambda$ in terms of $x$ and $y$
Since $\left( {0,0} \right)$ does not satisfy the constraint, we may assume that $x \ne 0$ and $y \ne 0$.
From Step 1, we obtain $\lambda = - x = \frac{y}{5}$. So $\lambda \ne 0$.
Step 3. Solve for $x$ and $y$ using the constraint
From Step 2, we obtain $y=-5x$. Substituting it in the constraint $g\left( {x,y} \right)$ gives
$5x - y - 4 = 0$
$5x + 5x = 4$
$x = \frac{2}{5}$
So, the critical point is $\left( {\frac{2}{5}, - 2} \right)$.
Step 4. Calculate the critical values
We evaluate $f$ at the critical point: $f\left( {\frac{2}{5}, - 2} \right) = - \frac{4}{5}$.
Next, we verify that $f$ has a minimum at the critical point $\left( {\frac{2}{5}, - 2} \right)$.
From the constraint $g\left( {x,y} \right) = 5x - y - 4 = 0$, we obtain $y = 5x - 4$.
Substituting $y = 5x - 4$ in $f$ gives
$g\left( x \right) = f\left( {x,y} \right) = x\left( {5x - 4} \right) = 5{x^2} - 4x$
As $x \to \infty $, we get $g \to \infty $. So, we conclude that the minimum value of $f\left( {x,y} \right) = xy$ subject to the constraint $5x - y = 4$ is $ - \frac{4}{5}$.
Method 2. setting $y = 5x - 4$ in $f\left( {x,y} \right)$
Substituting $y = 5x - 4$ in $f$ gives
$g\left( x \right) = f\left( {x,y} \right) = x\left( {5x - 4} \right) = 5{x^2} - 4x$
We find the critical points of $g$ by solving $g'\left( x \right) = 0$:
$g'\left( x \right) = 10x - 4 = 0$
So, the critical point is $x = \frac{2}{5}$.
Since $g{\rm{''}}\left( x \right) = 10 > 0$, we conclude that $g$ has a minimum at $x = \frac{2}{5}$. The minimum value is $g\left( {\frac{2}{5}} \right) = - \frac{4}{5}$.
Hence, we conclude that the minimum value of $f\left( {x,y} \right) = xy$ subject to the constraint $5x - y = 4$ is $ - \frac{4}{5}$.
