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Homework Help: In Dire need of assistance for optimisation (Area/volume)

  1. Sep 1, 2012 #1
    1. The problem statement, all variables and given/known data

    A square-based rectangular prism has a total surface area of 54cm^2. Determine the side lengths if the volume is a maximum and hence find the volume.

    I have done a number of these and I am getting really annoyed because I always get all the sides being equal no matter what I try. I have absolutely no clue what I am doing wrong but i need to know!!!

    2. Relevant equations

    SA=2l^2 + 4lh

    3. The attempt at a solution

    Ok So what i did was SA= 2l^2 +4lh=54 and rearranged for h so that h=(54-2l^2)/4l

    then I subbed into the volume equation to make it v=l^2((54-2l^2)/4l)=( 54l^2-2l^4)/4l=
    (54l-2l^3)/4= 54l/4-l^3/2, then i derived to make it 54/4 - 3l^2/2= 0. then i rearranged to find l... 3l^2/2=54/4, 3l^2/2= 13.5, l^2= 13.5/1.5, l^2=9, l=3. then i proved it was a max, i'm not going to show that here. anyway then i subbed l into h=(54-2/^2)/4l and ended up with 3 for h

    that means l and h are equal to 3 but that wouldn't make it a rectangular prism it would be a cube... what am i doing wrong? I am so frustrated by this!!!
  2. jcsd
  3. Sep 1, 2012 #2


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    Actually a cube is a special case of a rectangular prism. Squares are rectangles too! They're just special cases.

    If the answer weren't that it was a cube (because you can't allow it to be) then what you'll be doing is trying to find the maximum volume (which is when it's a cube) but not allowing [itex]l=h[/itex] which then means you'd need to take h to be very close to l as the answer, which is pretty silly.
  4. Sep 1, 2012 #3
    Hmm does that mean, I am correct? It is strange because every question that involves rectangles has been giving me trouble because I always seem to get an answer where the sides are equal! If I am correct in all of these cases then I have been tressing for nothing! Also may I ask you to assist me in answering a few business optimisation qustions! if so here is the one I am finiding most difficult:

    A manufacturer can sell x clock radios per week at d dollars per item, where x=600-3d.
    Production costs in dollars are C(x)=600+10x+1/2x^2.
    How should the clock radios be priced in order to maximise profits.

    I had no clue what to do, first i rearranged to isolate d(x) and then went from there but that did not work at all. I am confused about this type of question, also any others that ask how much we should sell something for.

    By the way thank you for helping with my first part you have been a great help and what you said was easy to understand :) not like a lot of others on here
  5. Sep 1, 2012 #4

    Ray Vickson

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    So, if you sell x radios per week at d dollars each, what is your weekly revenue? Now just express everything either in terms of x (using the relation between x and d to express d in terms of x), or else express everything in terms of d. Probably, expressing in terms of x is best because you still have to account for cost in order to finally compute profit = revenue - cost. Now you will have an expression for profit, P, in terms of number of clocks, x, and you can vary x until P is a maximum.

    BTW: it would be better to write 1/2 x^2 properly, as either 1/2 * x^2 or better still, as (1/2)x^2 or (1/2)*x^2. Doing that will make it clear that you do _not_ mean 1/(2x^2).

  6. Sep 1, 2012 #5
    I have no clue. Is it R(x)=600x-3dx? I really do not know!
  7. Sep 1, 2012 #6


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    Well, revenue =($ per item )(# of items)

    If you're not sure of your formula, check the units of the result:

    ($/item)(# of items) = # of dollars,

    so it works out.
  8. Sep 1, 2012 #7

    Ray Vickson

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    Of course you know; you are just not thinking it through! If I sell 5 units at $3 per unit, what is my revenue? If I sell 10,000 units at $7 per unit, what is my revenue? If I sell x units at d dollars per unit, what is my revenue?

  9. Sep 1, 2012 #8
    Summing a bit what we know and assuming [itex]x=q[/itex] and [itex]d=p[/itex] we have:

    [1] [itex]\hspace{1cm}[/itex] [itex]q=600-3p[/itex]

    [2] [itex]\hspace{1cm}[/itex] [itex]C(q)=600+10q+ \frac{1}{2} q^{2}[/itex]

    Now, we should add just for the sake of completeness that the problem is ill-posed as far as it doesn't explicitly say which kind of market are we facing. Indeed, you wrote that you should answer pointing out the price at which the profit is the highest, but that's a bit "monopolistic". This is relevant because it does change the formula to get the profit from, let's say, monopoly to perfect competition. In this case we assume that we are in perfect competition, because we lack a demand function for clocks and [1] is enough to make us think so. In particular let's remember that in perfect competition producers are price-takers. Then - as Ray Vickson adviced you - we have to focus on quantity.

    Then we have:

    [3] [itex]\hspace{1cm}[/itex] [itex]π(q)=R(q)-C(Q)[/itex]

    that in perfect competition means

    [4] [itex]\hspace{1cm}[/itex] [itex]π(q)=pq-C(Q)[/itex] [itex]\hspace{1cm}[/itex] with [itex]R(q)=pq[/itex]

    Now, here we are. You write [itex]R(x)=600x-3dx[/itex]. But again, as Ray Vickson pointed out, let's focus on [itex]q[/itex].

    Then, you have [itex]C(q)[/itex] and [itex]R(q)[/itex] as well. Indeed, we know from [4] what [itex]R(q)[/itex] is equal to in perfect competition. In particular, considering that we are focusing on [itex]π(q)[/itex], which is profit as function of quantity, we have to make [itex]p[/itex] disappear from the problem.
    Can we? Given [1], YES WE CAN (with some manipulations).

    [No, I am not from USA, so that's not a badly hidden political message! :biggrin:]
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