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Question: A camping supply company makes backpacks in two models, journey and trek. The journey model requires $4$ hours of labor and generates a profit of $40$ dollars. The trek model requires $6$ hours of labor and generates a profit of $80$ dollars. The company needs to profit at least $\$400$ per week. Their distributor will accept no more than $4$ trek backpacks and no more than $15$ journey backpacks per week. How many of each type of backpack should the company make to minimize the number of hours of labor?


My Attempt:

I let $x$ be the number journey and $y$ be the number of trek backpacks produced during the week. Thus, we have

$$40x+80y\geq 400\\0\leq x\leq 4\\ 0\leq y\leq 15\tag1$$

And an objective function of $4x+6y$. Thus, the minimized/maximized points are $(0,5),(0,15),(4,3),(4,15)$ and I got $0,5$ as the answer. Meaning $5$ treks and no journeys.

But the problem was wrong. It was supposed to be $(4,3)$.

Where did I go wrong?

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  • $\begingroup$ The objective function must be Min 4x + 6y, with 40x+80y>=400, 0<y<=4 and 0<=x<=15. You messed up the last two constraints with notation. $\endgroup$ – Satish Ramanathan Oct 31 '16 at 13:47
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You mix up the constraints. It should be $$ 0\le x\le 15,\quad 0\le y\le 4. $$

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