# Max/Min problem I'm fundamentally flawed in my understanding I think

• monet A
In summary, the conversation discusses frustration with maximum and minimum problems and a specific issue with finding the maximum volume of a closed box with a given surface area. The conversation then delves into a discussion about converting units, specifically the incorrect conversion of square meters to square centimeters. The conversation ends with some helpful tips for unit conversion.

#### monet A

I am entirely frustrated with maximum and minimum problems. I have had this issue with them since I first was introduced to them and I thought I had resolved it but it keeps getting up and biting me again.

The problem is I am not even sure what I am in the habit of doing wrong, so I can't fix it. Could somebody point out what I have done ar*eways here for me so I can get over this once and for all..?

A closed box is to be made with length equal to 3 times its width. The total surface area of the box will be 30m^2. Find the dimensions that give maximum volume in the box.

$$3000cm^2 = 2lw + 2hw + 2lh$$

$$\frac {3000 - 6w^2} {8w}} = h$$

$$V = l *w * h$$

$$V(w) = 3w * w * \frac {3000 - 6w^2} {8w}} = \frac {9000w - 18w^3} {8}}$$

$$V'(w) = \frac {9000 - 54w^2} {8}}$$

$$V''(w) = \frac {- 108w} {8}}$$

Now when I find the 0 value of dv/dw it is one value that can be positive or negative, naturally I disregard the negative value, but then plugging anything positive into the second derivative I will always get a negative slope, so all values >0 would return as a maximum. I can finish off the problem from there and get values.

However...

My problem is that if I work it out in units of metres by allowing 30m^2 to equal the surface area I get a value of ca 1.29 metres, and if I use cm^2 (above) as the surface area unit I get ca 12.9cm. Both work to give 30m^2 as the surface area but they are clearly not the same values yet they are working perfectly in the same equation?

Should I have found both of these as critical points when looking for 0 in my 1st derivative? If so, how, what did I miss...

please someone help me iron out this crinkle in my understanding because as you can see I am too confused to do it myself.

Last edited:
30 square meters is not equal to 3.000 square centimeters but 300.000 square centimeters.

$$1 \mbox{m}^2 = 10.000 \mbox{cm}^2$$

you have to reduce both dimensions by a factor of 100.

Galileo said:
30 square meters is not equal to 3.000 square centimeters but 300.000 square centimeters.

$$1 \mbox{m}^2 = 10.000 \mbox{cm}^2$$

you have to reduce both dimensions by a factor of 100.

Oh as in 100 cm * 100 cm = 10000 cm^2 I thought it was a fundamental flaw in my understanding but *that* fundamental is just embarrassing, thanks for pointing it out to me :yuck: I think
I am sure I won't let this happen again, I might get paranoid enough to question all my chemistry calculations though... what have I done?

don't worry, if you pay attention, you will see that people make this mistake all the time. pretty much any news article you read that converts units^2 does it incorrectly.

to make things worse even, in my 300 level geography class a few weeks ago, the instructor was saying something about how many square kilometers of rainforest were destroyed each year. I think it was like, 25,000km^2 were destroyed. he's like "anyone got a calculator, so we can put that in miles?". and this nerdly kid calculates it, and comes up with 14,000mi^2 (which is wrong, and dumb to boot, since he had a ti-8x with the conversion built in..). to make things worse, the instructor says "so, that's about 7,000miles by 7,000 miles."

yup, good thing I'm at a high quality university.

lol fishhy

monet: A good trick not to get confused about unit conversion is to do this: 1m = 100cm, so

$$30m^2 = 30 (100 cm)^2 = 30\cdot 100^2 \cdot cm^2 = 300000cm^2$$

cool huh?

So as soon as you know the "linear conversion", you can retrieve any other "higher degree" conversion. Another example, taken from physics: 1eV = 1.6*10^-19 J, so

$$5\cdot 10^9eV^3 = 5\cdot 10^9(1.6\cdot 10^{-19} J)^3 = 5\cdot 10^9\cdot (1.6\cdot 10^{-19})^3 J^3 = 8\cdot 10^{-10} J^3$$

## 1. What is a Max/Min problem in mathematics?

A Max/Min problem is a type of optimization problem in mathematics that involves finding the maximum or minimum value of a function within a given set of constraints.

## 2. How do you solve a Max/Min problem?

To solve a Max/Min problem, you need to first identify the objective function and the constraints. Then, you can use various techniques such as differentiation, substitution, or graphical methods to find the critical points. Finally, you evaluate the objective function at the critical points to determine the maximum or minimum value.

## 3. What is the difference between a local and global Max/Min?

A local Max/Min is the maximum or minimum value of a function within a specific interval or range. A global Max/Min, on the other hand, is the maximum or minimum value of a function in the entire domain.

## 4. Can you provide an example of a Max/Min problem?

One example of a Max/Min problem is finding the maximum area of a rectangle with a fixed perimeter. The objective function is A = lw (area = length x width), and the constraint is 2l + 2w = P (perimeter = 2 x length + 2 x width). By solving for l in the constraint equation and substituting it into the objective function, we can find the critical point and determine the maximum area.

## 5. What are some real-world applications of Max/Min problems?

Max/Min problems have various applications in real-world scenarios, such as in economics, engineering, and physics. For example, a company may use Max/Min problems to determine the optimal production level that maximizes profits or minimizes costs. In physics, Max/Min problems can be used to find the optimal trajectory of a projectile to reach a target.

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