Question

Maximize {eq}p = 2x + 3y

{/eq} subject to

{eq}3x + 8y leq 43

{/eq}

{eq}6x + 4y leq 44

{/eq}

{eq}x geq 0, y geq 0

{/eq}.

EXPERT ANSWER

The graph of the inequalities is

where we see the feasible region in purple. The vertices may be found from the equations

{eq}3x + 8y = 43 \

8y = 43 – 3x \

y = frac{43}{8} – frac{3}{8} x \

\

6x + 4y = 44 \

4y =