# Optimization problem/finding domain and proof

• miglo
In summary, the problem involves finding the dimensions of a metal box with a square base and no top that holds 1000 cubic centimeters and has the least production cost. The cost function is given as C(x) = x^2 + 0.004/x + 0.0002/x^2, with a derivative of C'(x) = 2x - 0.004/x^2 - 0.0004/x^3. The domain of the cost function needs to be determined and a proof needs to be provided for finding the minimum point.
miglo

## The Attempt at a Solution

so for this problem I've managed to find the function the gives the cost to make the box
after converting everything to the same units my function is $$C(x)=x^2+\frac{.004}{x}+\frac{.0002}{x^2}$$
so the derivative for the cost function is $$C'(x)=2x-\frac{.004}{x^2}-\frac{.0002}{x^3}$$
setting this equal to zero and using maple, i get $x=0.1494530180$ plus 3 more zeros but one is negative and the other two are complex so in the context of this problem those 3 make no sense
after plotting my function i can see the critical point i have found is in fact the minimum, but I am also asked to find the domain of my cost function and write up a proof that i have found a minimum and this is where I am stuck.

## Homework Equations

if the cost function was given to me as just a regular function, one not in the context of this problem then i would say my domain would be all real except x=0, but I don't think it'll work for this

for my proof, i think ill have to use intermediate value theorem and extreme value theorem? but the extreme value theorem only works for closed intervals, and i still haven't found the domain

## The Attempt at a Solution

I don't know your units, but when I use b = base of box and h = height of box = 1000/b (h, b in cm) and express all costs in $I get (using Maple 9.5): $$\displaystyle C(b) = \frac{2}{5} + \frac{b^2}{10000} + \frac{2}{b} + \frac{400}{b^2}$$ Here, the material cost is 10^(-4) ($/cm^2) and the seaming cost is 5*10^(-4) $/cm. Putting b = 100*x gives the cost $$\displaystyle C(x) = x^2 + 0.4 + \frac{0.2}{x} + \frac{.04}{x^2}$$ The derivative C'(x) has the form N(x)/(50*x^3), where N(x) = 100*x^4-x-4. The function N(x) is strictly increasing if x > (1/400)^(1/3), and N is negative at that point. Thus, there is exactly one positive root of N(x) = 0. RGV Last edited: well the problem states that the metal box can only hold up to 1000 cm^3, therefore the volume should be 1000 cm^3 but the cost for the material are all stated in meters and not centimeters, so i converted from centimeters to meters 100cm=1m (100cm)^3=(1m)^3 1,000,000cm^3=1m^3 (1m^3/1,000,000cm^3)*(1000cm^3/1)=1000m^3/1,000,000=0.001m^3 therefore the volume of the metal box in cubic meters is 0.001 but i also know the volume of a box is length*width*height, and in this problem i have length and width being equal since my base is a square, so naming the length and width of my sides x and my height y, i get V=x^2y but V=0.001=x^2y, so x^2y=0.001 solving for y i get y=0.001/x^2 so my cost function, C(x)=x^2+4x(0.001/x^2)+4*0.05(0.001/x^2) so C(x)= x^2+0.004/x+0.0002/x^2 did i do something wrong here to get my cost function? and yeah i just realized my derivative is wrong, it should be C'(x)=2x-0.004/x^2-0.0004/x^3 however this is pretty much the easy part for me, what I am having trouble with is finding the domain of C(x), and proving that i have found a minimum miglo said: well the problem states that the metal box can only hold up to 1000 cm^3, therefore the volume should be 1000 cm^3 but the cost for the material are all stated in meters and not centimeters, so i converted from centimeters to meters 100cm=1m (100cm)^3=(1m)^3 1,000,000cm^3=1m^3 (1m^3/1,000,000cm^3)*(1000cm^3/1)=1000m^3/1,000,000=0.001m^3 therefore the volume of the metal box in cubic meters is 0.001 but i also know the volume of a box is length*width*height, and in this problem i have length and width being equal since my base is a square, so naming the length and width of my sides x and my height y, i get V=x^2y but V=0.001=x^2y, so x^2y=0.001 solving for y i get y=0.001/x^2 so my cost function, C(x)=x^2+4x(0.001/x^2)+4*0.05(0.001/x^2) so C(x)= x^2+0.004/x+0.0002/x^2 did i do something wrong here to get my cost function? Sorry, I totally misread the question. I thought it was 5$ a meter instead of 5 cents a meter.

and yeah i just realized my derivative is wrong, it should be C'(x)=2x-0.004/x^2-0.0004/x^3
however this is pretty much the easy part for me, what I am having trouble with is finding the domain of C(x),
[/QUOTE]

Well, in what points in C not defined? Can I evaluate C(2), C(3.02)? (yes). Can I evaluate C(x) for all x? (No)

and proving that i have found a minimum

Prove that your function is decreasing before the "minimum" and increasing after. So prove that your derivative is negative before the "minimum" and positive after.

Or you can calculate the second derivative C''(x) and prove that it is >0 at the "minimum".

miglo said:
well the problem states that the metal box can only hold up to 1000 cm^3, therefore the volume should be 1000 cm^3
but the cost for the material are all stated in meters and not centimeters, so i converted from centimeters to meters

100cm=1m
(100cm)^3=(1m)^3
1,000,000cm^3=1m^3
(1m^3/1,000,000cm^3)*(1000cm^3/1)=1000m^3/1,000,000=0.001m^3

therefore the volume of the metal box in cubic meters is 0.001

but i also know the volume of a box is length*width*height, and in this problem i have length and width being equal since my base is a square, so naming the length and width of my sides x and my height y, i get V=x^2y
but V=0.001=x^2y, so x^2y=0.001 solving for y i get y=0.001/x^2
so my cost function, C(x)=x^2+4x(0.001/x^2)+4*0.05(0.001/x^2)
so C(x)= x^2+0.004/x+0.0002/x^2

did i do something wrong here to get my cost function?

and yeah i just realized my derivative is wrong, it should be C'(x)=2x-0.004/x^2-0.0004/x^3
however this is pretty much the easy part for me, what I am having trouble with is finding the domain of C(x), and proving that i have found a minimum

Your C(x) is wrong. My previous post was in error, and I have corrected it below. The area of the material (from which you cut out 4 corners of size y) is length * width = (x + 2y)^2 = x^2 + 4*y^2 + 4*x*y. You have y = 0.001/x^2, so the cost is C(x) = x^2 + 4*(0.001/x^2)^2 + 4*x*0.001/x^2 + 4*(.05)*0.001/x^2. This will have terms in x^2, 1/x, 1/x^2 and 1/x^4. Therefore, when you put it over a common denominator, the derivative of C(x) will have a polynomial of degree 6 in the numerator, so there will generally be 6 roots. However, plotting shows there is only one real, positive root.

RGV

@micromass
well then my domain will be all reals except for x=0 right?

and for the proof could i just use the intermediate value theorem on C'(x) in the interval [0.1,0.2] since C'(0.1)<0 and C'(0.2)>0 then there exists a zero in the interval, namely the x value i found earlier, then since C'(x) goes from negative to positive, that means C(x) decreases then increases after

miglo said:
@micromass
well then my domain will be all reals except for x=0 right?

Right!

and for the proof could i just use the intermediate value theorem on C'(x) in the interval [0.1,0.2] since C'(0.1)<0 and C'(0.2)>0 then there exists a zero in the interval, namely the x value i found earlier, then since C'(x) goes from negative to positive, that means C(x) decreases then increases after

sounds good!

## 1. What is an optimization problem?

An optimization problem is a mathematical problem that involves finding the maximum or minimum value of a given function, while satisfying a set of constraints. The goal is to find the "optimal" solution that gives the best possible outcome.

## 2. How do you find the domain of a function?

To find the domain of a function, you need to identify all the possible values of the independent variable (x) that the function can take. This means looking for any restrictions or limitations on x, such as values that would result in division by zero or negative values inside a square root.

## 3. Why is it important to find the domain of a function?

Finding the domain of a function is important because it tells us the set of all possible input values for the function. This helps us to understand the behavior of the function and also to avoid any undefined values or errors when evaluating the function.

## 4. What is a proof in optimization?

In optimization, a proof is a logical argument or a series of steps that demonstrate the validity of a solution to an optimization problem. It shows that the solution is indeed the optimal value for the given problem and satisfies all the necessary conditions.

## 5. How do you know if a solution to an optimization problem is valid?

A solution to an optimization problem is considered valid if it satisfies all the constraints and gives the best possible outcome. This can be verified by checking if the solution satisfies all the necessary conditions, such as the first or second derivative test for finding the maximum or minimum value of a function.

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