Answer:
see below
Step-by-step explanation:
If we let X represent the number of bagels produced, and Y the number of croissants, then we want ...
(a) Max Profit = 20X +30Y
(b) Subject to ...
6X +3Y ≤ 6600 . . . . . . available flour
X + Y ≤ 1400 . . . . . . . . available yeast
2X +4Y ≤ 4800 . . . . . . available sugar
_____
Production of 400 bagels and 1000 croissants will produce a maximum profit of $380.
__
In the attached graph, we have shaded the areas that are NOT part of the solution set. (X and Y less than 0 are also not part of the solution set, but are left unshaded.) This approach can sometimes make the solution space easier to understand, since it is white.
The vertex of the solution space that moves the profit function farthest from the origin is the one we are seeking. The point that does that is (X, Y) = (400, 1000).
Given: C(N) = 15,000 + 8000N <span>
In the above equation simply substitute:
N(t) = 100t - 5t^2
for N
</span>
<span>Therefore:
C(t) = 15,000 + 8000{ 100t-5t^2 }
C(t) =15,000 + 800,000t - 40,000t^2.</span>
at t = 5
C(5) = 15,000 + 800,000*5
- 40,000*(5)^2
<span>C(5) = 3,015,000</span>
There is no solution for the first one
if you eliminate y you get 2 equations
-13x - 13z = -25
-13x - 13x = -15
- there is no solution to theses
a23 means the element in the second row and the 3rd column
so its -5
Answer: C
Step-by-step explanation:
The question asks what was eaten so on the first day, 1 part of three, and on the second day another portion of 3 so 2/3 was eaten, aka C
Answer: 61, 62, 63
Step-by-step explanation:
61 + 62 + 63 = 186