Hello!
We can write this as the system of equations below. We have boxes A, B, and C.
A+B+C=9
B=A+1
C=B+1
First of all we can plug our B and C values into the first equation and solve for A.
A+A+1+A+1+1=9
3a+3=9
3a=6
a=2
Now we can plug our a value into the other equations.
B=2+1
B=3
C=3+1
C=4
Therefore, there are 2 rocks in one box, three in another, and 4 in the last.
I hope this helps!
y - 3
g(y) = ------------------
y^2 - 3y + 9
To find the c. v., we must differentiate this function g(y) and set the derivative equal to zero:
(y^2 - 3y + 9)(1) - (y - 3)(2y - 3)
g '(y) = --------------------------------------------
(y^2 - 3y + 9)^2
Note carefully: The denom. has no real roots, so division by zero is not going to be an issue here.
Simplifying the denominator of the derivative,
(y^2 - 3y + 9)(1) - (y - 3)(2y - 3) => y^2 - 3y + 9 - [2y^2 - 3y - 6y + 9], or
-y^2 + 6y
Setting this result = to 0 produces the equation y(-y + 6) = 0, so
y = 0 and y = 6. These are your critical values. You may or may not have max or min at one or the other.
