A rug cleaning company sells three models. EZ model weighs 10 pounds, packed in a 10 -cubic-foot box. Mini model weighs 20 pounds, packed in an 8-cubic-foot box. Hefty model weighs 60 pounds, packed in a 28-cubic-foot box. A delivery van has 336 cubic feet of space and can hold a maximum of 480 pounds. To be fully loaded, how many of each should it carry if the driver wants the maximum number of Hefty models?
EZ models Mini models Hefty models

Step 1 ：Let's denote the number of EZ, Mini, and Hefty models as $x$, $y$, and $z$ respectively.

Step 2 ：The total weight of the models in the van should not exceed 480 pounds. So, we have the inequality: $10x+20y+60z\le 480$.

Step 3 ：The total volume of the models in the van should not exceed 336 cubic feet. So, we have the inequality: $10x+8y+28z\le 336$.

Step 4 ：Since the driver wants the maximum number of Hefty models, we need to maximize $z$.

Step 5 ：From the weight inequality, we can express $x$ in terms of $y$ and $z$: $x=48-2y-6z$.

Step 6 ：Substitute $x$ into the volume inequality: $10(48-2y-6z)+8y+28z\le 336$.

Step 7 ：Simplify the inequality to get: $z\le 4-\frac{y}{7}$.

Step 8 ：Since $z$ is an integer, the maximum value of $z$ is 4 when $y=0$.

Step 9 ：Substitute $z=4$ and $y=0$ into the equation for $x$ to get: $x=48-2(0)-6(4)=24$.

Step 10 ：So, the van should carry 24 EZ models, 0 Mini models, and 4 Hefty models to be fully loaded and have the maximum number of Hefty models.