Devin runs a factory that makes Blu-ray players. Each R80 takes 9 ounces of plastic and 2 ounces of metal. Each G150 requires 3 ounces of plastic and 6 ounces of metal. The factory has 270 ounces of plastic, 348 ounces of metal available, with a maximum of 18 R80 that can be built each week. If each R80 generates $8 in profit, and each G150 generates $9, how many of each of the Blu-ray players should Devin have the factory make each week to make the most profit? R80:

G150:

Best profit:

9514 1404 393

  R80: 12

  G150: 54

  Best Profit: $582

Step-by-step explanation:

Let x and y represent the numbers of R80 and G150 players, respectively. The constraints of the problem are ...

  0 ≤ x ≤ 18 . . . . . a maximum of 18 R80 can be built

  0 ≤ y . . . . . . . . . only non-negative numbers can be built

  9x +3y ≤ 270 . . . . ounces of plastic used cannot exceed 270

  2x +6y ≤ 348 . . . . ounces of metal used cannot exceed 348

The objective is to maximize the profit function ...

  P(x, y) = 8x +9y

The attached graph shows profit is a maximum of $582 per week when 12 R80 players and 54 G150 players are produced.

_____

Since the maximum profit is at a value of x less than 18, we didn't bother to graph that constraint.

hello! just here for points! though i wish i knew!

27/4x^6 y^8

step-by-step explanation:

4(3x^2y^4)^3 (2x^3y^5)^4

=

108x^6 y^12 /16x^12 y^20

=

108 /16x^6 y^8

=

27 /4x^6 y^8