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
Answer:
13.1
Step-by-step explanation:
Add all the number together and divide by the number of students (aka total number of responses reported)
16+13+19+16+8+10+10=92
92/7= 13.1428 or about 13.1
Domain is the x and range is y so your domain is 2,3,4,5 and your range is 0 and 4
3.9 can be rounded to 4.
5.3 can be rounded to 5.
4 x 5 = 20
That tells you that 3.9 x 5.3 has a digit in the tens place.
Then you can proceed to multiply 39 and 53, which equals 2067.
(Note: I removed the decimals on purpose)
Now we know that in 2067, the 2 should be in the tens place. That means 0 is in the ones place and that 6 is in the tenths place. That leaves 7 to be in the hundreths place.
Therefore, 3.9 x 5.3 = 20.67
