Answer:
5 units
Step-by-step explanation:
Let point O be the point of intersection of the kite diagonals.
|OF| = 2, |OH| = 5
|FH| = |OF| + |OH| = 2 + 5 = 7
FH and EG are the diagonals of the kite. Hence the area of thee kite is:
Area of kite EFGH = (FH * EG) / 2
Substituting:
35 = (7 * |EG|) / 2
|EG| * 7 = 70
|EG| = 10 units
The longer diagonal of a kite bisects the shorter one, therefore |GO| = |EO| = 10 / 2 = 5 units
x = |GO| = |EO| = 5 units
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
