Determine the maximum value of the objective function, p. p= 15x+12y 6x+4y≤2000 2x+4y≤1000 x≥0 y≥0

Maximum (250,125) Answer

Step-by-step explanation:

This is more of a graphing problem than it is anything else.

Begin by graphing all 4 given equations.

When you do that, mark the intersection points of at least 2 lines. In this case it is exactly 2 lines for each intersecting point.

6x + 4y <= 2000 and 2x + 4y <= 1000 intersect at (250,125)

x=>0 and y=>0 intersect at (0,0)

6x + 4y <=2000 and x => 0 intersect at 333.333

2x + 4y <=1000 and y>=0 intersect at  0,250.

Any other intersection points fall outside the range of the givens. The shaded part we are interested in is sort of a very dark green/blue. It is the interior of the quadrilateral determined by the 4 vertices that are marked.

Now all you have to do is determine the maximum point using P=15x + 12y

For 0,0                 P = 15*0 + 12,0 = 0

For 0,250            P = 15*0 + 12*250 = 3000

For 333.3333,0   P = 15*333.3333 + 12*0 = 5000 rounded.

For 250,125        P = 15*250 + 12*125 = 5250  Which is the maximum

Answer (250,125) produces the maximum value Answer

the expression (1.26)2t reveals the approximate rate of increases in the number of plants semiannually.

hope this : )

b

step-by-step explanation: