At an Oregon fiber-manufacturing facility,
an analyst estimates that
the weekly number of pounds
of acetate fibers that
can be produced is given
by the function z equals f of x comma y,
where z equals the weekly number
of pounds of acetate fiber,
x equals the number of
skilled workers at the plant,
and y equals the number of
unskilled workers at the plant.
We have several questions here.
The first question asks us
to find the weekly number
of pounds of fiber that can be produced
with 11 skilled workers
and 36 unskilled workers.
So for this question, we want
to evaluate the given function
when x equals 11 and y equals 36.
So we'd have f of 11 comma 36 equals
12,500 times 11
plus 4,000 times 36
plus seven times 11 squared
times 36 minus five times 11 cubed.
To save some times I've
already evaluated this,
it comes out to
305,337.
And this would be the
number of weekly pounds.
Given the plant has 11 skilled workers
and 36 unskilled workers.
Next, we're asked to find an expression
for the rate of change of the output
with respect to the
number of skilled workers.
Remember the number of
skilled workers is equal to x,
which means you want to find
the partial derivative of f
with respect to x.
To do this we'll differentiate
with respect to x,
treating y as a constant.
So the partial derivative
of f with respect to x
would be equal to, well
the derivative of 12,500x
with respect to x would be 12,500
plus the derivative of 4,000y
with respect to x would be zero.
Again, because we're
treating y as a constant.
Plus the derivative of seven x squared y
with respect to x, we
would multiply by two,
that would give us 14.
Subtract one from the exponent of x,
so we'd have 14xy.
And the derivative of
negative five x to the third
with respect to x would
be negative 15x squared.
So we'd have minus 15x squared.
Number three, we want to
find the rate of change
of the output with
respect to skilled workers
when 11 skilled workers
and 36 unskilled workers are employed.
Because we're trying to
find the rate of change
of the output with
respect to skilled workers
or with respect to x, we want to evaluate
the partial derivative with respect to x
when x equals 11 and y equals 36.
So the partial with
respect to x at 11 comma 36
would be equal to 12,500
plus 14 times x which is 11
times y which is 36
minus 15 times x squared
which would be 11 squared.
Again, I've already
determined this amount.
It comes out to
16,229.
And the units on this would be
weekly pounds per skilled worker.
And because this is the
rate of change of the output
with respect to skilled workers,
the units would be weekly
pounds per skilled worker.
Next, we're asked to find an expression
for the rate of change of
the output with respect to
the number of unskilled workers.
And again, because y equals the
number of unskilled workers,
we now want to find the
partial derivative of f
with respect to y.
To do this we'll differentiate
with respect to y,
treating x as a constant.
So the partial with respect to y
would be equal to the
derivative of this first term
with respect to y would be zero,
because we're treating x as a constant.
The derivative of 4,000y with
respect to y would be 4,000.
The derivative of seven x
squared y with respect to y
would be seven x squared,
because the derivative of y equals one.
And the derivative of
negative five x cubed
with respect to y would be zero.
So here's our partial with respect to y.
And our last question we're asked to find
the rate of change of
the output with respect
to unskilled workers
when 11 skilled workers
and 36 unskilled workers are employed.
So because we're measuring
the change in output
with respect to the unskilled workers now,
we want to evaluate the partial
derivative with respect to y
when x equals 11 and y equals 36.
So the partial with
respect to y at 11 comma 36
is equal to 4,000 plus
seven times x squared.
But x is 11 so 11 squared.
Again, I've already determined this value.
It comes out to 4,847.
And because this measures
the change in output
with respect to unskilled workers,
the units would be weekly
pounds per unskilled worker.
So I think this problem
is a nice application
of why partial derivatives are useful.
I hope you found this helpful.
