Maximum Revenue Question

Stellar Steel can produce x tonnes of regular-quality steel and y tonnes of super-quality steel per day, where y=40-5x/10-x. The market price for regular quality steel is half the market price for super quality steel. How many tonnes of regular quality steel should the company produce to maximum revenue.

The answer to this problem is 5.53 tonnes.

How do I go about solving this problem?

I found the derivative to the equation however, I'm not sure what the derivative helps me to solve?

Can someone lead me through the steps to solve this equation? Thanks (:

let R(x) equal the revenue. So R(x) = 2((40-5x)/(10-x)) + x
Of course you would multiply R(x) times the $/ton of regular steel.

d/dx(2((40-5x)/(10-x)) + x) = (2((10-x)(-5) - (40-5x)(-1))/(10-x)^2) + 1

R ' (x) = (-20)/ (x^2 - 20x + 100) + 1 You said you know how to use the quotient rule.

Now, we want to set R ' (x) = 0. This is where the original equation R (x) in maximized.

R ' (x) = 0 = ((-20)/(x^2 - 20x + 100)) + 1

((-20)/(x^2 - 20x + 100)) = -1

Now multiply both sides by the quadratic denominator:

-20 = - (x^2 - 20x + 100)

-20 = x^2 + 20x - 100 All this I did - just to get rid of the quotient

-x^2 + 20x - 80 = 0 Again, we must set the derivative function to zero, because this is where the slope is zero of the revenue. Where the slope of the revenue is zero, is where the maximum revenue is. This whole process is all about getting more money.

Next, is the quadratic equation: x = (-20 + sqrt(80)/-2

x = 5.527864045 tons of steel

Company wants to maximise revenue and it is that eqauation for which you need to find derivate and not the given one.

Let us say the price of super quality steel is $S. This means that price for regular steel must be S/2.