Answer:
1. Objective function is a maximum at (16,0), Z = 4x+4y = 4(16) + 4(0) = 64
2. Objective function is at a maximum at (5,3), Z=3x+2y=3(5)+2(3)=21
Step-by-step explanation:
1. Maximize: P = 4x +4y
Subject to: 2x + y ≤ 20
x + 2y ≤ 16
x, y ≥ 0
Plot the constraints and the objective function Z, or P=4x+4y)
Push the objective function to the limit permitted by the feasible region to find the maximum.
Answer: Objective function is a maximum at (16,0),
Z = 4x+4y = 4(16) + 4(0) = 64
2. Maximize P = 3x + 2y
Subject to x + y ≤ 8
2x + y ≤ 13
x ≥ 0, y ≥ 0
Plot the constraints and the objective function Z, or P=3x+2y.
Push the objective function to the limit in the increase + direction permitted by the feasible region to find the maximum intersection.
Answer: Objective function is at a maximum at (5,3),
Z = 3x+2y = 3(5)+2(3) = 21
20 is the answer your welcome you will get it right don’t worry
Answer:
7^2 ( little 2 in the corner)
1) We can determine by the table of values whether a function is a quadratic one by considering this example:
x | y 1st difference 2nd difference
0 0 3 -0 = 3 7-3 = 4
1 3 10 -3 = 7 11 -7 = 4
2 10 21 -10 =11 15 -11 = 4
3 21 36-21 = 15 19-5 = 4
4 36 55-36= 19
5 55
2) Let's subtract the values of y this way:
3 -0 = 3
10 -3 = 7
21 -10 = 11
36 -21 = 15
55 -36 = 19
Now let's subtract the differences we've just found:
7 -3 = 4
11-7 = 4
15-11 = 4
19-15 = 4
So, if the "second difference" is constant (same result) then it means we have a quadratic function just by analyzing the table.
3) Hence, we can determine if this is a quadratic relation calculating the second difference of the y-values if the second difference yields the same value. The graph must be a parabola and the highest coefficient must be 2