An open-top rectangular box is being constructed to hold a volume of 350 cubic inches.
The base of the box is made from material costing 6 cents per square inch. The front of the box must
be decorated and will cost 12 cents per square inch. The remainder of the sides will cost
2 cents per square inch. Find the dimensions that will minimize the cost of constructing this box.
Let's first diagram the box, where the dimensions are x by y by z. Because the volume must be
350 cubic inches, we have a constraint that x times y times z must equal 350.
Now, before we talk about our cost function, let's talk about the surface area of the box.
Because the top is open, we only have five faces. Let's find the area of the five faces that would
make up the surface area. Notice the area of the front face would be x times z, which should
also be the same as the area in the back, so the surface area has two xz terms. Notice the right
side of the right face would have an area of y times z, which would be the same as the left,
so the surface area contains two yz terms. Finally, the bottom has an area of x times y,
and because the top is open, we only have one xy term in the surface area.
Now we'll convert the surface area to the cost equation.
The bottom costs six cents per square inch, where the area of the bottom is
xy. Notice how for the cost function, we multiply the xy term by six cents.
Because the front costs 12 cents per square inch, where the area of the front would be xz,
we'll multiply this xz term by 12 cents in the cost function.
The remaining sides cost two cents per square inch, so these three areas are all multiplied
by zero point zero two or two cents. Combining like terms, we have this cost function here.
But notice how we have three unknowns in this equation,
so now we'll use a constraint to form a cost equation with two variables.
If we solve our constraint for x by dividing both sides by yz,
we can make a substitution for x into our cost function, or we can substitute this fraction
here for x here and here. If we do this, we get this equation here, and if we simplify,
notice how a factor of z simplifies out and here a factor of y simplifies out.
So for this first term, if we find this product and then move the y up,
we would have 49y to the negative one, and then for the last term,
if we found this product and moved the z up, we'd have plus 21z to the negative one.
Now our goal is to minimize this cost function. For the next step, we'll find the critical points.
Critical points are where the function is going to have max or min function values,
and they occur where the first order partial derivatives are both equal to zero
or where either does not exist. Then once we find the critical points,
we'll determine whether we have a max or min value using our second order partial derivatives.
So on this slide, we're finding both the first order and second order partial derivatives.
We have to be a little careful here though because our function is a function of y and z,
not x and y like we're used to. For the first partial with respect to y, we would differentiate
with respect to y, treating z as a constant, which would give us this partial derivative here.
For the first partial with respect to z, we would differentiate with respect to z
and treat y as a constant, which would give us this first order partial derivative.
Now, using these first order partial derivatives, we can find these second order
partial derivatives. To find the second order partial with respect to y, we would differentiate
this partial derivative with respect to y again, giving us this. The second partial with respect
to z would be found by differentiating this partial derivative with respect to z again,
giving us this. Notice how it's given using a negative exponent and in fraction form. Finally,
for the mixed partial or the second order partial with respect to y and z,
we would differentiate this partial with respect to z, which notice how it would just give us 0.04.
So now we're going to set the first order partial derivatives equal to zero
and solve as a system of equations. Here are the first order partials set equal to zero.
This is a fairly involved system of equations, which we'll solve using substitution.
I decided to solve the first equation here for z. So I added this term to both sides of the equation
and then divided by 0.04, giving us this value here for z. But if we find this quotient and
move y to the negative two to the denominator, we can also write z as this fraction here.
Now that we know z is equal to this fraction, we can substitute this for z into the second equation
here, which is what we see here. But notice how this is raised to the exponent of negative two,
so this would be one thousand two hundred twenty-five to the negative two
divided by y to the negative four. We can take the reciprocal, which would give us y to the fourth
divided by one million five hundred thousand six hundred twenty-five, and here's the twenty-one.
Now that we have an equation with just one variable y, we want to solve this for y. For the
first step, there is a common factor of y, so y equals zero would satisfy this equation, and would
be a critical point. But we know we're not going to have a dimension of zero, so we'll just ignore
that value and set this expression here equal to zero and solve, which is what we see here.
So we're going to isolate the y cubed term and then cube root both sides of the equation.
If we add this fraction to both sides of the equation and then change the order
of the equation, this is what we would have. Now from here, to isolate y cubed,
we have to multiply by the reciprocal of this fraction here. Notice how the
left side simplifies to just y cubed, and this product here is approximately this value here.
Now to solve for y, we would cube root both sides of the equation or raise both sides of
the equation to the one-third power, and this gives us y is approximately 14.1918.
Now to find the z coordinate of the critical point, we can use this equation here where
z is equal to 1225 divided by y squared, which gives us z is approximately 6.0822.
We don't need it right now, but I went ahead and found the corresponding x value as well
using our volume formula to solve for x, so x would be approximately 4.0548.
Because we only have one critical point, we can probably assume this point is going
to minimize the cost function. But to verify this, we'll go ahead and use the critical point
and the second order partial derivatives just to make sure. We'll use this formula here for d and
the values of the second order partial derivatives to determine whether we have a relative max or min
at this critical point when y is approximately 14.19 and z is approximately 6.08.
Here are the second order partials that we found earlier.
We'll be substituting this value for y and this value for z into these second order partials.
We should be a little careful though because remember we have a function of y and z,
not x and y like we normally would, so these x's would be these y's and these y's would be the z's.
So the second ordered partial with respect to y is here, the second order partial with respect to z
is here, and here's the mixed partial squared. Notice how it comes out to a positive value.
So if d is positive and so is the second partial with respect to y, looking at our notes here,
that means we have a relative minimum at our critical point.
Therefore, these are the dimensions that would minimize the cost of our box. This was the
x coordinate from the previous slide, here's the y coordinate, and here's the z coordinate,
which again are the dimensions of our box. So the front width would be x,
which is approximately 4.05 inches, the depth would be y, which is approximately 14.19 inches,
and the height would be z, which is approximately 6.08 inches.
Let's finish by looking at our cost function where we have the cost function
in terms of y and z in three dimensions. This would be the surface
where these lower axes would be the y and z axes, and the cost would be along the vertical axis.
We can see there's a low point here, and that occurred at a critical point that we found.
I hope you found this helpful.