Glorious Gadgets is a retailer
of astronomy equipment.
They purchase equipment from a supplier and then sell it
to customers in their store.
The functions C of x given here models our total inventory
costs in dollars, as a function of x the lot size
for each of their orders from the supplier.
The inventory includes such things as purchasing,
processing, shipping and storing the equipment.
We have two questions to answer.
One, what lot size should Glorious Gadgets order
to minimize their total inventory costs?
This answer must be a whole number.
And two, what is their minimum total inventory cost?
So we need to minimize this cost function,
so we'll apply the first derivative test,
which means the first step is to find the derivative,
set it equal to zero, and find the critical numbers,
which is where the min or max values will occur.
So if this is C of x, notice that C prime of x
would be equal to the derivative of three point five x,
that's three point five,
plus the derivative of 9,000 x to the negative one.
So we multiply by negative one.
That would be 9,000 x raised to the power of
negative one minus one, that's negative two.
And the derivative of a constant is zero,
so now we'll set this equal to zero
and solve for x.
Let's go and add 9,000 x to the power of negative two
to both sides.
So we're going to have 9,000 x to the power of negative two
equals three point five.
Let's try both of these in fraction form.
So we'd have 9,000 over x squared equals
three point five over one.
And now that we have a proportion, we can cross multiply
and solve for x.
Three point five times x squared must equal 9,000 times one.
So three point five x squared equals 9,000.
Now isolate x squared by dividing both sides
by three point five.
Simplifying, we have x squared is equal to 9,000
divided by three point five.
And now we'll take the square root of both sides
of the equation and solve for x.
Now, algebraically, we are going to have a positive
and negative solution here.
Because we know x cannot be negative,
we'll only find the positive solution
or the principle square root.
So now we'll go to the calculator.
So we have x equals the square root of 9,000
divided by three point five.
Again, our answer must be a whole number,
so we'll say our critical number is approximately
x equals 51.
So we have one critical number, which is
x is approximately 51.
And we can't assume this is the x value
that's going to minimize the inventory.
We do need to apply the first derivative test.
So to do this, let's sketch a number line.
Our critical number is approximately 51,
so we'll use the first derivative test,
which means we'll test the sum of the first derivative
to the left and right of 51 to see if the function
is increasing or decreasing.
From this information, we can determine
whether we have a min or max value at x equals 51.
So for our test values, let's use, say, x equals 40,
and x equals, let's say, 60.
Any value to the right and left of 51 would work.
And, again, we'll sub these test values into
our first derivative which we found here.
So going back to the calculator,
C prime of 40 would be three point five
minus 9,000
times 40
raised to the power of negative two.
Notice how the first derivative is negative,
which means the function is decreasing
on this interval.
So in this interval, the first derivative was negative
or less than zero,
which means the function is decreasing on this interval.
And now we'll test the sum of the first derivative
at x equals 60.
So we because we already have this expression in here,
we can just press sec and enter
and change the 40 to a 60.
And notice how when x equals 60,
the first derivative is positive.
So if the first derivative is positive over this interval,
which means the function is increasing over this interval.
And this is good news because that means that
x equals 51 or approximately 51, the function changes
from decreasing to increasing,
and therefore we are going to have a minimum
function value at x equals 51.
So to answer the question, a lot size
of 51
will minimize the total inventory cost.
Now for the second question,
what is the minimum total inventory cost,
we want to evaluate the cost function,
not the derivative function, at x equals 51.
Let's do this on the next slide.
So the minimum total inventory cost is going to be
C of 51, which is going to be equal to three point five
times 51
plus 9,000
times 51
raised to the negative one plus 6,000.
And now we'll go back to the calculator.
So we have three point five times 51
plus 9,000
times 51
raised to the power of negative one,
right arrow, plus 6,000.
Because these are dollars, we round to the nearest cent.
So we'll say it's approximately $6,354.97.
So the minimum total inventory cost is $6,354.97.
Notice how we use the whole number solution
to the first question to answer the second question.
Before we go, let's go ahead and verify the result
graphically, and we'll do this by analyzing the graph
of C of x.
So our cost function C of x is graphed here,
and we found, using the first derivative test,
that we had a minimum function value of x equals 51,
which is approximately here, at this low point.
And we can tell from the graph that the total cost
of the inventory,
when x equals 51,
is around $6,300.
Notice how we can also verify the function is decreasing
to the left of 51 and increasing to the right of 51.
Even though it's not increasing very fast,
to the right of 51 it is going uphill,
moving toward the right.
I hope you found this helpful.