anyone help me this ?

Question

Josh runs a factory that makes DVD players. Each T50 takes 8 ounces of plastic and 3 ounces of metal. Each FS20 requires 4 ounces of plastic and 6 ounces of metal. The factory has 304 ounces of plastic, 348 ounces of metal available, with a maximum of 20 T50 that can be built each week. If each T50 generates $8 in profit, and each FS20 generates $11, how many of each of the DVD players should Josh have the factory make each week to make the most profit?

T50:  

FS20:  

Best profit:  

Answer 1

Let x = the number of T50s built and y = the number of FS20s built. The inequalities are:

Plastic: 8x + 4y ≤ 304

Metal: 3x + 6y ≤ 342

Production: 0 ≤ x ≤ 20, 0 ≤ y

The equation to be optimized (Profit) is:

Profit = $8x + $11y

1. Graph the inequalities (you can do it by hand or use a graphing app)
2. Identify the Feasible Zone
3. identify the (x,y) coordinates of each vertex of the Feasible Zome
4. Plug each vertex (x,y) value into the profit equation to find the one yielding the max profit. The corresponding x and y values are the number of T50s and FS20s.

If you have any questions, ask them in the comments.