Continuing from the setup in the question linked above (and using the same symbols/variables), we have
The next part of the question asks to maximize this result - our target function which we'll call
- subject to
.
We can see that
is quadratic in
, so let's complete the square.
Since
are non-negative, it stands to reason that the total product will be maximized if
vanishes because
is a parabola with its vertex (a maximum) at (5, 25). Setting
, it's clear that the maximum of
will then be attained when
are largest, so the largest flux will be attained at
, which gives a flux of 10,800.
Answer:
Option C
Step-by-step explanation:
You forgot to attach the expression that models the cost of the camping trip during the three days. However, by analyzing the units, the answer can be reached.
The total cost has to be in units of $.
There are two types of costs in the problem:
Those that depend on the number of days ($/day
)
Those that depend on the number of students and the number of days ($/(student * day))
If there are 3 days of camping and b students, then you have to multiply the costs that depend on the days by the number of days (3), and the costs that depend on the number of students you have to multiply them by 'b'
So, if the costs that must be multiplied by 'b' are only those that depend on the number of students, the coefficient of b must be:
3 days (Cost of training + Cost of food Miscellaneous expenses :).
Therefore the correct answer is option C:
C. It is the total cost of 3 days per student of Mr. Brown, with training, food and miscellaneous expenses.
The expression that represents the total expense should have a formula similar to this:
![y = (3 days) *([\frac{20.dollars}{(day * student)} + \frac{30.dollars}{(student * day)} + \frac{50.dollars}{(student * day)}] b + \frac{200}{day}) + 1050.dollars](https://tex.z-dn.net/?f=y%20%3D%20%283%20days%29%20%2A%28%5B%5Cfrac%7B20.dollars%7D%7B%28day%20%2A%20student%29%7D%20%2B%20%5Cfrac%7B30.dollars%7D%7B%28student%20%2A%20day%29%7D%20%2B%20%5Cfrac%7B50.dollars%7D%7B%28student%20%2A%20day%29%7D%5D%20b%20%2B%20%5Cfrac%7B200%7D%7Bday%7D%29%20%2B%201050.dollars)
y = 3 ($ 100b + $ 200) + $ 1050
A=Value of investment.
A=12000(1+0.015)^2
A= <span>£12362.70</span>
Answer: £23,360
Step-by-step explanation:
You can use algebra to solve this.
Assume that the original investment is x.
In the beginning of 2015 the balance had increased by 2.5% which is denoted as:
= 1.025x
Carol then withdrew a thousand:
= 1.025x - 1,000
This value then increased by 2.5% again by the beginning of 2016:
= 1.025 * (1.025x - 1,000)
Relevant expression therefore is:
1.025 *(1.025x - 1,000) = 23,517.60
1.025x - 1,000 = 23,517.60/1.025
1.025x = 22,944 + 1,000
x = 23,944/1.025
x = £23,360
Length is 2 and width is 3
2 X 3 is 6
