A farmer wishes to employ apple pickers during harvest time I'm having trouble again, this time with the word problem below.
Now, about all I really have is an equation for part B and a guess as to the answer for part A, and I'm not sure either is correct.
For part C, I think I need to take $C_p$ and $C_h$ and set them to zero in order to find the values of $p$ and $h$ (as attempted below), but I'm not sure.

The Questions 

A farmer wishes to employ apple pickers during harvest time. Each picker can harvest $625$ apples per hour and is paid $\$6$ per hour. In addition, the farmer must pay a supervisor $\$10$ per hour and the union $\$10$ for each picker employed. Finally, if $v$ apples are picked, then a service charge of $\$\frac{50,000}{\sqrt{v}}$ is levied against the farmer.
a. Briefly explain why the levy charge of $\$\frac{50,000}{\sqrt{v}}$ motivates the farmer to hire an adequate number of pickers.
b. Set up a function which represents the farmer's total cost.  
c. Find the number of apple pickers which should be employed to minimize the farmer's cost.  
d. Find the number of apples which would be picked at this minimum cost.


My Answers
a. The levy charge motivates the farmer to hire an adequate number of pickers because as the number of apples picked increases, the amount of the service charge decreases.
b. Where $p$ is the number of pickers and $h$ is the number of hours,
$$C = 6ph + 10h + 10p + \frac{50,000}{\sqrt{v}}$$
$$= 6ph + 10h + 10p +  \frac{50,000}{\sqrt{625ph}}$$
$$= 6ph + 10h + 10p +  \frac{2,000}{\sqrt{ph}}$$
c.
$$C_p = 6h + 10 - \frac{1,000}{p^{3/2}\sqrt{h}} = 0$$
$$10 = \frac{1,000}{p^{3/2}\sqrt{h}} - 6h$$
$$C_h = 6p + 10 - \frac{1,000}{h^{3/2}\sqrt{p}} = 0$$
$$10 = \frac{1,000}{h^{3/2}\sqrt{p}} - 6p$$
$$\frac{1,000}{p^{3/2}\sqrt{h}} - 6h= \frac{1,000}{h^{3/2}\sqrt{p}} - 6p$$
$$\frac{1,000}{p^{3/2}\sqrt{h}} - \frac{1,000}{h^{3/2}\sqrt{p}} = 6h - 6p$$
And from here, I'm lost.
 A: You have two equations in two unknowns
$$10 = \frac{1,000}{h^{3/2}\sqrt{p}} - 6p\\\frac{1,000}{p^{3/2}\sqrt{h}} - \frac{1,000}{h^{3/2}\sqrt{p}} = 6h - 6p$$
If you combine terms in the second, you get $$\frac {1000(h-p)}{(ph)^{3/2}}=6(h-p)\\6(ph)^{3/2}(h-p)={1000}(h-p)$$
If you assume $h \neq p$ you can use this to substitute into the first and you get $10=0$, so you must have $h=p$.  Then plugging into the first gives $6p^3+10p^2=1000$, with solution $p=5=h$.
A: It looks good so far. For part c, you should find the minimum value of the function, which will not be when $p=0$ and $h=0$, because you'd be dividing the last term of the function by zero.
Once you know the minimum value of the function, you will know how many pickers and how many hours they work, thus you can find the number of apples picked for part d.
A: I don't understand what you are asking but:
If you want to minimize $C(h,p) =  C_h + C_p $ just do:
$$
\nabla C(h,p) = 0
$$
A: You seem on the right course.  You know your possible minimum is when both partial derivatives are zero. I think you're making the algebra too tough tough.  When you set Cp = 0 you have an equation of two variables. Similarly when you set Ch =0 you have an equation of two variables.  Try to state p in terms of h in each equation.  It's going to be ugly. 
It may be that you need to revisit your cost function in B. How does it look if you get the radical out of the denominator? How do the partial derivatives look?
