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The constraints of a problem are listed below. What are the vertices of the feasible region? 4x+3y<124x+3y\lt 12, 2x+6y<152x+6y\lt15, x>0x\gt 0, y>0y\gt 0.

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Given:

4x+3y<122x+6y<15x>0y>0\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:

4x+3y=122x+6y=15x=0y=0\begin{align*} 4x+3y&=12 \\ 2x+6y&=15 \\ x&=0 \\ y&=0 \end{align*}

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

We note that (3,0)(3,0) and (0,4)(0,4) are two points on the line 4x+3y=124x+3y=12 (as they satisfy the equation) and thus we can draw the line 4x+3y=124x+3y=12 by drawing the two points and drawing a straight line through the pair of points.

We note that (1.5,2)(1.5,2) and (2.5,0)(2.5,0) are two points on the line 2x+6y=152x+6y=15 (as they satisfy the equation) and thus we can draw the line 2x+6y=152x+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<124x+3y<12 is the area on the side of the line 4x+3y=124x+3y=12 that contains the origin (as x=0x=0 and y=0y=0 satisfies the inequality). Similarly, the area 2x+6y<152x+6y<15 is the area on the side of the line 2x+6y=152x+6y=15 that contains the origin (as x=0x=0 and y=0y=0 satisfies the inequality).

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

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