## Related questions with answers

The constraints of a problem are listed below. What are the vertices of the feasible region? $4x+3y\lt 12$, $2x+6y\lt15$, $x\gt 0$, $y\gt 0$.

Solution

VerifiedGiven:

$\begin{align*} 4x+3y&<12 \\ 2x+6y&<15 \\ x&>0 \\ y&>0 \end{align*}$

We will start by drawing the four corresponding lines in the plane:

$\begin{align*} 4x+3y&=12 \\ 2x+6y&=15 \\ x&=0 \\ y&=0 \end{align*}$

$x=0$ represents the vertical axis and $y=0$ represents the horizontal axis.

We note that $(3,0)$ and $(0,4)$ are two points on the line $4x+3y=12$ (as they satisfy the equation) and thus we can draw the line $4x+3y=12$ by drawing the two points and drawing a straight line through the pair of points.

We note that $(1.5,2)$ and $(2.5,0)$ are two points on the line $2x+6y=15$ (as they satisfy the equation) and thus we can draw the line $2x+6y=15$ by drawing the two points and drawing a straight line through the pair of points.

Next, we note that the area $4x+3y<12$ is the area on the side of the line $4x+3y=12$ that contains the origin (as $x=0$ and $y=0$ satisfies the inequality). Similarly, the area $2x+6y<15$ is the area on the side of the line $2x+6y=15$ that contains the origin (as $x=0$ and $y=0$ satisfies the inequality).

The feasible region is then the intersection of these two regions that lies above the $x$-axis and to the right of the $y$-axis.

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