# Where did I go wrong in this linear programming problem?

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?

• 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. – Satish Ramanathan Oct 31 '16 at 13:47

You mix up the constraints. It should be $$0\le x\le 15,\quad 0\le y\le 4.$$