Answer:
$180000
Step-by-step explanation:
Let's c be the number of chair and d be the number of desks.
The constraint functions:
- Unit of wood available 4d + 3c <= 2000 or d <= 500 - 0.75c
- Number of chairs being at least twice of desks c >= 2d or d <= 0.5c
c >= 0
d >= 0
The objective function is to maximize the profit function
P (c,d) = 400d + 250c
We draw the 2 constraint functions (500 - 0.75c and 0.5c) on a c-d coordinates (witch c being the horizontal axis and d being the vertical axis) and find the intersection point 0.5c = 500 - 0.75c
1.25c = 500
c = 400 and d = 0.5c = 200 so P(400, 200) = $250*400 + $400*200 = $180,000
The 500 - 0.75c intersect with c-axis at d = 0 and c = 500 / 0.75 = 666 and P(666,0) = 666*250 = $166,500
So based on the available zones in the chart we can conclude that the maximum profit we can get is $180000
B.C. counts down to 0, and that's when A.D. starts counting up. That means you can add 582 and 1643, to get their difference. Pythagoras and Newton were born 2,225 years apart.
A Rhombus has four equal straight sides. We can assume that the rhombus in this problem is square-shaped with 90° angles.
m1 = 18x ; 90 = 18x ; 90/18 = x ; 5 = x
m2 = x + y ; 90 = 5 + y ; 90 - 5 = y ; 85 = y
m3 = 30z ; 90 = 30z ; 90/30 = z ; 3 = z
x = 5 ; y = 85 ; z = 3
x + y + z ⇒ 5 + 85 + 3 = 93
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