Question

(Optimization problem, please work through it with me!)

Suppose an airline policy states that all baggage must be box-shaped with a sum of length, width, and height not exceeding 108 in. What are the dimensions and volume of a square-based box with the greatest volume under these conditions?

Answer 1

The constraint is L + W + H = 108 right? And the objective function is V = LWH? I don't know where to go from there...

Answer 2

Ok so since it is square based that means that I should reformat the constraint to:
2L + H = 108 and the objective function to V = L^2H

Answer 3

yep, exactly.

Answer 4

But to be honest, I don't read from the text what they want maximized, minimized, the volume?

Answer 5

ahh greatest volume!, sorry, it's late here.

Answer 6

yeah no problem! So next do I solve for L or H? and plug it into the Volume function?

Answer 7

yes, always make sure that your substitution doesn't make the function complicated, for single variable calculus you want the function to be only dependent on one variable.

With simple I mean, always match it to the level of differentiation you are at.

Answer 8

in this case it doesn't really matter, I am just saying it.

Answer 9

Ok so in the future I will be able to solve it with multiple variables? But for now the optimization problems will need to be reduced to a single variable?

Answer 10

yes, they use La Grange multipliers there, don't worry about that just yet. But keep in mind that you always try to express your function in terms of only one variable. For example, this would be such a function:

\[ V= f(L)=L^2(108-2L)=108L^2-2L^3 \]

Answer 11

you see, this function represents the volume, but it only depends on the Length, on nothing else.

Answer 12

so now I solve for critical points? and how do I know the limit? is it (0, 54) ?

Answer 13

not limit i mean domain

Answer 14

\[ f'(x)=0 \] is one condition for it to be an extrema (Maximum Minimum)

\[f''(x)<0 \] means that it is a maximum.

Answer 15

well the domain of the function we have written is \(\mathbb{R}\). But for our solutions, only positive numbers make sense, and zero shouldn't be an answer too.

Answer 16

\[f'(x)=216L-6L^2=0\]

\[ f'(x) = 6L(36-L)=0 \]

Answer 17

wait so we first find roots and then we find critical points? or do roots not matter?

Answer 18

excuse me, I am from Switzerland and a lot of the common terms you have don't ring a bell for me. I don't know what you mean with critical points, we actually don't find roots of the function, we find the roots of the derivative which show you where the function takes the highest or lowest point.

Here is the graph of the function:

http://www.wolframalpha.com/input/?i=108x%5E2-2x%5E3

We are interested in the points where this function has it's highest value (Maximum)

Answer 19

ohhh ok. Critical point means when slope is zero, what you said

Answer 20

ahhh, okay, we call that extrema here, or just extreme. But I am glad to know that, thank you (-:

Answer 21

So i got V'(L) = 216L - 6L^2 when I solved for the derivative

Answer 22

yup and you want this slope to be equal to zero.

Answer 23

so when L = 36, the slope is zero

Answer 24

yes.

Answer 25

Now you have basically all the results you need, but make sure, depending on how rigorous your Professor/Teacher is, always make sure that you check with the second derivative what kind of critical point that is.

Answer 26

In my class they taught us to now plug in this result (L = 36 in our case) to the original function along with the end points of the function (the domain points). And in our case we find the highest value

Answer 27

but i am not sure what the domain is for this function

Answer 28

Hmm okay, I think I understand, but to be honest and a bit bold - it makes less sense here to me. Because this Function is defined everywhere, for our purpose we just care about the numbers which are larger than 0 and smaller than infinity.

Answer 29

So you see, you have nothing to compare it with.

Answer 30

yes but the sum has to be less than or equal to 108 right?

Answer 31

Haven't you done second derivative yet to proof that a point is a Maximum or a Minimum? You know that there are two kind of critical points, one is the maximum, where the function takes it's highest values and the one where it's the lowest.

Answer 32

yeah but what do I compare it with? I know I will have to do f''(36) but what else to make sure?

Answer 33

yes that's true, the sum has to be less or equal to 108, so you can use that as a constraint, but it's not a real constraint by the function.

Answer 34

f''(36)<0 you must check that, if that's the case, the point you have found is a maximum and you are done in my opinion.

Answer 35

ok cool. I guess I just want to be more secure of my answer in case there are other maxima, minima

Answer 36

In this case, you don't have any other options to check, only if for example your result would have been x=72

72+72=144>108, so this would have been not a correct answer.

Answer 37

\[f''(x)=216-12L \\
f''(36)=-216<0 \longrightarrow \text{Max}\]

Answer 38

should read as the second derivative, obviously this latex script isn't working too well in here.

Answer 39

ok so v(36) is the maximum!

Answer 40

and I can check again to see if it is a maximum by plugging it into the second derivative, which gives me a negative, which means concave down

Answer 41

yes it means that it's a maximum, concave down.

Answer 42

Thank you have a good day! I might have more questions later for other problems.