rate of change -profits

[rajendrarama]

the profit P made by an entertainment centre when selling x bags of lollies was modelled by the equation

P = 2.5x - (1/20000)x^2 -3000 0<=x<= 50000

a) For what values of 'x' is the centre making a positive profit?
b) for what values of 'x' is the profit (i) increasing? (ii) decreasing?

I could do,

dP/dx = 2.5 - (x/10000)
b) increasing would be when (x/10000) <= 2.5 , it means when -3000<=x<=25000
decreasing would be when (x/10000) >=2.5, it means when 25000<=x<=50,000

could you please confirm the above, and I am not sure how to solve part a),
any ideas?

 

[stapel]

the profit P made by an entertainment centre when selling x bags of lollies was modelled by the equation

P = 2.5x - (1/20000)x^2 -3000 0<=x<= 50000

a) For what values of 'x' is the centre making a positive profit?
b) for what values of 'x' is the profit (i) increasing? (ii) decreasing?

I could do,

dP/dx = 2.5 - (x/10000)
b) increasing would be when (x/10000) <= 2.5 , it means when -3000<=x<=25000
decreasing would be when (x/10000) >=2.5, it means when 25000<=x<=50,000

could you please confirm the above...?

I haven't checked your math, but your method looks fine.

I am not sure how to solve part a), any ideas?

Do the same sort of thing as you did for (b): Create the appropriate inequality (using the profit function, P(x)), and solve for the interval(s) on which P(x) is positive.

 

[HallsofIvy]

If you know about derivatives, you should certainly know about continuous functions. And important properties of continuous functions are
1) they are positive (or negative) on intervals and
2) those intervals are separated by points where the function value is 0.

So: find all values of x where P(x)= 0, then decide on which of the intervals determined by those points the function is positive.

(In fact, since the given P(x) is quadratic you don't really need to use Calculus at all- just complete the square.)