We're asked to find the minimum cost
of a rectangular box of volume, 190 cubic centimeters,
who's top and bottom cost 4 cents per square centimeter
and whose sides cost 8 cents per square centimeter.
To model the information,
we will use the box shown here on the right.
The dimensions are X centimeters,
by Y centimeters, by Z centimeters.
To begin, we know the volume
is equal to 190 cubic centimeters,
and therefore V equals X, Y, Z equals 190.
This is the constraint of the problem.
And now set up the cost equation.
We know the top and bottom
costs 4 cents per square centimeter.
Looking at the diagram,
notice how the bottom is X centimeters by Y centimeters,
and so is the top.
And therefore the total area
of the top and bottom is two times X times Y
and the cost is 4 cents per square centimeter,
which means your total costs
for the top and bottom
would be four times 2XY.
And now let's focus on the front and the back.
The front has dimensions X centimeters by Z centimeters,
and so does the back.
The total area of the front
and back would be two times X times Z
at a cost of 8 cents per square centimeter.
And therefore the cost of the front
and back would be eight times two XZ,
and then finally for the right and left sides,
the dimensions are Y by Z and therefore
the total area of the right and left sides
would be two times Y time Z
at a cost of 8 cents per square centimeter giving
us eight times two YZ
for the cost of the right and left sides.
And now we want to minimize the cost function,
given the constraint of the volume,
we will do this using the method of Lagrange Multipliers,
and therefore we need to write
the cost function as F of X comma Y, comma Z,
and the constraint in the form
of g of X comma Y, comma Z equals zero.
For the cost function.
We have F of X comma Y, comma Z
equals 8XY, plus 16XZ, plus 16YZ.
To form g, we set the volume equation equal
to zero by subtracting 190 on both sides,
which gives us g of X comma Y, comma Z
equals XYZ minus 190.
And now to solve using the Lagrange multipliers,
we need to set up the system of equations shown
here below using the partial derivatives
as well as the constraint.
Our first equation is the partial of F
with respect to X equals Lambda times
a partial of g with respect to X.
To find the partial of F with respect to X,
we differentiate with respect to X treating
both Y and Z as constants.
The partial of f with respect to X is 8Y
plus 16Z, plus zero or just 8Y plus 16Z.
This is equal to Lambda times
a partial of g with respect to X.
To find the partial with respect to X,
we differentiate g with respect
to X treating Y and Z as constants,
which gives us YZ minus zero or just YZ.
The second equation is the partial F
with respect to Y equals Lambda times partial of g
with respect to Y.
To find the partial of f with respect to Y,
we differentiate g with respect to Y
treating both X and Z as constants.
Which gives us 8X plus zero, plus 16Z
or 8X plus 16Z equals Lambda times
the partial g with respect to Y, which is just XZ.
Third equation is the partial of f
with respect to Z equals Lambda times partial
of g with respect to Z.
For the partial of f with respect to Z,
we differentiate with respect to Z
treating X and Y as constants,
which gives us zero plus 16X plus 16Y
or 16X plus 16Y,
and then we have equals Lambda times
the partial of g with respect to Z,
which is just XY.
And our fourth equation is g of X
comma Y, comma Z equals zero,
which in our case is XYZ minus 190 equals zero.
And now we need to solve this system of equations,
which is where things get a little involved,
thus high number of the equations to keep things organized.
So we have our system of equations here at the top.
For the first step
we're going to solve equation one for Y
and equation two for X.
Once we do this, hopefully we'll see
a relationship between X and Y.
So starting with the equation one to solve
for Y, we need to have the terms
with the factor of Y on the same side.
So we subtract Lambda YZ
on both sides and also subtract 16Z on both sides,
then factor out Y on the left
and then divide both sides by eight minus Lambda Z,
which gives us Y equals negative 16Z
divided by the quantity eight minus Lambda Z.
And now we go to equation two
and solve for X.
We get the terms with the factor
of X on the same side,
by subtracting Lambda XZ on both sides.
And also subtracting 16Z.
Next, factor X from the left side, and then divide
by the quantity eight minus Lambda Z,
which gives us X equals negative 16Z divided
by the quantity eight minus Lambda Z.
And if we take a look at the expressions
for X and Y,
notice how both X and Y are equal
to negative 16Z divided by the quantity eight
minus Lambda Z.
And therefore we can conclude that X is equal to Y.
So the next step we're going
to solve equation two for Z
and equation three for Y,
and hopefully see a relationship
between Z and Y.
So starting with the equation two,
we now solve for Z,
which means we subtract Lambda XZ
on both sides and also subtract 8X on both sides,
which gives us a second equation.
Next, we factor out Z on the left
and then divide by the quantity 16
minus Lambda X,
which gives us Z equals negative 8X
divided by the quantity of 16 minus Lambda X.
Going to equation three,
we want to solve for Y.
First step, get the terms with the factor of Y
on the same side.
So we subtract Lambda XY on both sides,
and also subtract 16X on both sides.
Next, factor out the common factor
of Y on the left, and then divide
by the quantity 16 minus Lambda X on both sides,
which gives us Y equals to negative 16X
divided by the quantity 16 minus Lambda X.
And now if we compare the expressions
for both Z and Y,
notice how denominators are the same,
but the numerators are not.
The numerator for Z is negative eight X.
The numerator for Y is negative 16 X,
but also notice how negative 8X
times two does give us a negative 16X,
which means we multiply Z by two,
it would be equal to Y.
Or we can say Y is equal to 2Z.
And we also know X is equal to Y.
So we could also say X is equal to 2Z.
Now that we have relationships among X, Y, and Z,
we can solve for X, Y, and Z using
an equation for the constraint.
Let's do this on the next slide.
Again, we just found that X
is equal to Y,
and also, Y is equal to 2Z
which is also equal to X.
If we wanted to solve for Z,
notice how we would divide by two.
We can also say Z is equal
to one half Y as well as one half X.
So now we'll go to equation four,
let's go and write equation four
as X, Y, Z equals 190
and now substitute X for Y
and one half X for Z,
so we have an equation in one variable.
We would have X times X times Z
is one half X, equals 190.
Multiplying on the left,
we have one half X cubed equals 190.
To solve for Xqb, multiply both sides by two,
which gives us X cubed equals 380.
And now to solve for X,
we take the cube root of both sides of the equation.
And now we know that X
is equal to the cube root of 380.
We also know X equals Y.
So we also know Y is equal
to the cube root of 380.
And we also know Z is equal
to one half X of one half Y.
And therefore Z is equal
to one half times cube root of 381
or for cube root of 380 divided by two.
And of course, all of these units would be centimeters.
So we just found the dimensions of the box,
where the volume is 190 cubic centimeters,
and the cost is minimized.
But the question here is
to find the minimum cost.
So now we take these values of X, Y, and Z,
and substitute them back into
our cost function or our cost equation given here.
I write down this to save some time.
I have substituted the cube root of 380
for both X and Y, and the cube root of 380
divided by two for Z.
looking at the second and third products,
notice 16 divided by two is eight here,
as well as here,
which means we can write this product
as 24 times the square of the cube root of 380.
And now we go to the calculator
and remember, because we set up the cost equation,
and since this cost will be in cents,
which gives us approximately 1259 cents.
Let's go ahead and convert this
to dollars and cents by dividing by 100,
because there's 100 cents in a dollar.
The minimum cost is approximately $12 and 59 cents.
I hope you found this helpful.