Partial derivative is derivative over only 1 variable.
here we have 2 variables so it depends on which variable you want to make partial derivative.
case 1.
df/dx = -3y
case 2
df/dy = 5y^4 - 3x
Answer:
The value of q that maximize the profit is q=200 units
Step-by-step explanation:
we know that
The profit is equal to the revenue minus the cost
we have
---> the revenue
---> the cost
The profit P(q) is equal to

substitute the given values



This is a vertical parabola open downward (because the leading coefficient is negative)
The vertex represent a maximum
The x-coordinate of the vertex represent the value of q that maximize the profit
The y-coordinate of the vertex represent the maximum profit
using a graphing tool
Graph the quadratic equation
The vertex is the point (200,-120)
see the attached figure
therefore
The value of q that maximize the profit is q=200 units
Answer:
atleast 52
Step-by-step explanation:
Given that an employment agency requires applicants average at least 70% on a battery of four job skills tests.
An applicant scored 70%, 77%, and 81% on the first three exams,
Since weightages are not given we can assume all exams have equal weights
Let x be the score on the 4th test
Then total of all 4 exams = 
Average should exceed 70%
i.e.
Comparing the two totals we have

Hemust score on the fourth test a score atleast 52 to maintain a 70% or better average.
Answer:
There was a 25% increase.
Step-by-step explanation:
The fact that the first chosen student was a girl is already decided, so there is no need to find the probability of that happening. There are a total of 13+10, or 23 students. A girl was already selected, so that leaves 22 students. The chance that a boy is chosen is 10/22, or 5/11.