# Math Help - Max and Min values question.

1. ## Max and Min values question.

A local furniture manufacturer sells cedar pato sets. The company can sell x units each month at a price of p(x)=1000-x in dollars, where the cost, C, of producing x units per day is, C(x)=3000+19x^2.

Determine a) the price that will maximize the profits
and b) the break even point.

Not sure where to start...

2. Originally Posted by Harne
A local furniture manufacturer sells cedar pato sets. The company can sell x units each month at a price of p(x)=1000-x in dollars, where the cost, C, of producing x units per day is, C(x)=3000+19x^2.

Determine a) the price that will maximize the profits
and b) the break even point.

Not sure where to start...
Lets see, we sell at $1000 - x$ BUT the cost to produce it is $3000 + 19x^2$

So wouldn't the profit be $1000 - x - (3000 + 19x^2)$

So find the derivative of x in the profit formula and set x into the price formula.

3. Originally Posted by janvdl
Lets see, we sell at $1000 - x$ BUT the cost to produce it is $3000 + 19x^2$

So wouldn't the profit be $1000 - x - (3000 + 19x^2)$

So find the derivative of x in the profit formula and set x into the price formula.
No the sales are $1000-x$ units per day when the price per unit is $x$, so
the revenue is $(1000-x)x$, and the cost to produce $x$ units is $3000+19x^2$.

So the profit is:

$1000x - x^2 -3000 - 19x^2 = -20x^2+1000x -3000$.

RonL

4. Originally Posted by CaptainBlack
No the sales are $1000-x$ units per day when the price per unit is $x$, so
the revenue is $(1000-x)x$, and the cost to produce $x$ units is $3000+19x^2$.

So the profit is:

$1000x - x^2 -3000 - 19x^2 = -20x^2+1000x -3000$.

RonL
Ah ok, didnt realise that i should have multiplied with an extra $x$