We want to solve the following problem
using Calculus techniques.
We going to determine the number of units, X,
needed to be made and sold to maximize the profit,
which is P of X in dollars,
given the revenue and cost functions.
We also want to know what is the maximum profit.
So the first thing we have to remember here
is that the profit
is going to be equal to
the revenue function
minus the cost function.
So, P of X
is going to be equal to, with the revenue function,
which is 60X
minus 0.5 x squared
minus the cost function,
which is the quantity 3x plus eight.
Remember, it's important that we put the cost function
in parentheses so that we subtract this entire function,
not just the 3x.
Let's go and simplify this.
Here we can just drop the parentheses.
Here, we want to subtract both terms
or think of distributing a negative one.
So we have minus 3x
minus eight
Notice how we have 2 X terms.
Let's go ahead and combine them.
I'm also going to put the terms in descending order.
So I'll put the squared term first
and then 60x minus 3x
is 57x.
So plus 57x
minus eight.
Now to review, if you want to maximize the profit
using calculus techniques,
you want to first determine
their critical numbers, which occur where the derivative
of this function is either undefined or equal to zero.
So now let's find the derivative.
P prime of X
is going to be equal to, here,
we'll apply the power, so multiply by 2,
and then subtract one from the exponent,
that's just going to be negative one x or just negative x.
The derivative of 57x would just be 57.
The derivative of negative eight would be zero.
Notice how this derivative is never going to be undefined.
So let's set it equal to zero and solve for x.
This equation here is pretty straight forward to solve.
We would add x to both sides of the equation.
We would have 57 equals x.
Now I want to take a moment and go back
and take a look at this function here.
Notice if we were to graph this,
this would be parabola that opened down
because A) the coefficient of x squared is negative.
So if we have a parabola that opens down
and we're trying to
maximize this function, this high point here
would represent the maximum and notice that this point here,
the vertex, this slope with a tangent line would be zero,
which means the derivative
would be equal to zero at this high point.
The reason this is helpful is we now know that x equals 57.
We do have the maximum profit.
But remember, x is the number of units.
So to maximize the profit
then we need to make and sell 57 items
or units.
Now I'm going to determine the maximum profit,
which occurs when x equals 57.
So the maximum profit would be equal to P of 57.
So using one of these forms of the profit function,
we'll substitute 57 for x.
Let's go ahead and use this form here.
So we have negative 0.5 times 57 squared
plus 57 times 57
and then minus eight.
And I've already evaluated this to save some time.
This comes out to
One thousand six hundred sixteen dollars
and 50 cents
and now we've answered both questions.
I hope you found this helpful.