Re: Inequality word problem!

Let x be the number of A grade tea packages, and let y be the number of B grade packages. We want to maximize

$\displaystyle 1.5x + 2y$

subject to the constraints

$\displaystyle \frac{x}{3} + \frac{y}{2} \le 45$

$\displaystyle \frac{2x}{3} + \frac{y}{2} \le 70$

This can be solved by drawing the latter two inequalities on a coordinate plane and using a "linear programming" technique.

Re: Inequality word problem!

Quote:

Originally Posted by

**richard1234** Let x be the number of A grade tea packages, and let y be the number of B grade packages. We want to maximize

$\displaystyle 1.5x + 2y$

subject to the constraints

$\displaystyle \frac{x}{3} + \frac{y}{2} \le 45$

$\displaystyle \frac{2x}{3} + \frac{y}{2} \le 70$

This can be solved by drawing the latter two inequalities on a coordinate plane and using a "linear programming" technique.

You will also need to draw the inequalities $\displaystyle \displaystyle \begin{align*} x \geq 0 \end{align*}$ and $\displaystyle \displaystyle \begin{align*} y \geq 0 \end{align*}$.

Re: Inequality word problem!

So about the 1.5x+2y part, how would i maximize that? Do i graph that on the same plane as well somehow?

And to find the maximum profit, I choose the point where the two lines intersect right?

Re: Inequality word problem!

Quote:

Originally Posted by

**Calcgirl** So about the 1.5x+2y part, how would i maximize that? Do i graph that on the same plane as well somehow?

And to find the maximum profit, I choose the point where the two lines intersect right?

The maximum value will be at one of the corner points of your feasible region (that you get when you graph the inequalities). So substitute each of these points into your objective function to see which gives the maximum.