Example (film) | Wiskunde op Tilburg University

We solve the following extremum problem by the use of the first-order condition.

$\dfrac{z'_x(x,y)}{z'_y(x,y)}=\dfrac{g'_x(x,y)}{g'_y(x,y)}\Leftrightarrow \dfrac{2y}{2x+3} =\dfrac{4}{1}$ . Hence,

$y=4x+6$ .

We plug this into the restriction:

$4x+y=10$ then becomes

$4x+4x+6=10$ . Solving for

$x$ gives

$x=\frac{1}{2}$ .

Hence,

$y=4\cdot \frac{1}{2}+6=8$ .

This gives

$z(\frac{1}{2},8)=32$ . We still have to check whether this is a maximum. We can do this by determining two feasible combinations of

$x$ and

$y$ ; one with a smaller

$x$ -value (and hence a larger

$y$ value) and one with a larger

$x$ -value (and hence a smaller

$y$ -value).

Since

$z(1,6)=30<32=z(\frac{1}{2},8)$ and

$z(\frac{1}{4},9)=31\frac{1}{2}<32=z(\frac{1}{2},8)$ it holds that

$z(\frac{1}{2},8)=32$ is a maximum.