Precalculus (6th Edition) Blitzer

Published by Pearson
ISBN 10: 0-13446-914-3
ISBN 13: 978-0-13446-914-0

Chapter 7 - Section 7.6 - Linear Programming - Exercise Set - Page 873: 19

Answer

300 cartons of food and 200 cartons of clothing.
1583700104

Work Step by Step

Step 1. Assume $x$ cartons of food and $y$ cartons of clothing are shipped. Step 2. Based on the given conditions, the number of people helped can be written as $z(x,y)=12x+5y$ Step 3. We can convert the constraints into inequalities as $\begin{cases} x\geq0, y\geq0\\50x+20y\leq19,000\\20x+10y\leq8000 \end{cases}$ Step 4. Graphing the above inequalities, we can find the vertices and the solution region as a four-sided area in the first quadrant. Step 5. With the objective equation and vertices, we have $z(0,0)=12(0)+5(0)=0$, $z(0,800)=12(0)+5(800)=4000$, $z(300,200)=12(300)+5(200)=4600$, $z(380,0)=12(380)+5(0)=4560$ Step 6. Based on the above results, we can find the maximum number of people helped as $z(300,200)= 4600$ with 300 cartons of food and 200 cartons of clothing.
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