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September 28

How to calculate maximum volume of a box made with a 7x9 inch paper.

For a project, we are to calculate the maximum volume of a box that is made with a 7x9 piece of paper. The box is made by cutting squares off the corners and folding it. Help? —Preceding unsigned comment added by 67.34.27.6 (talk) 00:27, 28 September 2008 (UTC)[reply]

Does the box has to be a perfect cube? Or can it be any shape? 122.107.132.245 (talk) 00:41, 28 September 2008 (UTC)[reply]

Obviously it cannot be a perfect cube if it's 7×9 and the cuts out of the corners are squares.

If you cut x-by-x-inch squares out of each corner, then the box is x inches deep. You've then cut x from each end of the 9-inch long side, so what's left of the length is 9 − 2x, and similarly 7 − 2x remains of the width. Multiply to get the volume. (You haven't told us what number x is, that is how big the squares are that you cut out of the corners.) Michael Hardy (talk) 00:48, 28 September 2008 (UTC)[reply]

From first principles and using Back-of-the-envelope_calculation. We have

SurfaceArea = 7 * 9 = 63

Surface area for any arbitary box (with sides a,b and c) is

BoxSurfaceArea = 2 * (a*b + a*c + b*c)

BoxVolume = a*b*c

Simple manipulation get you

63/2 = (a*b + a*c + b*c)

Next assume that box is a cube, (aka a=b=c)

63/2 = 3 * a^2

a = Sqrt(63/6)

CubeVolume = (63/6)^(3/2) = 34.024

Next we check the wikipedia page on the cube and find out that "A cube has the largest volume among cuboids (rectangular boxes) with a given surface area."

Thus the maximum theoretical volume is 34.024.

122.107.132.245 (talk) 01:18, 28 September 2008 (UTC)[reply]

This "maximum theoretical volume" ignores the fact that the shape cannot be a cube, and that some of the paper will be cut away. I already gave the full solution above. Michael Hardy (talk) 03:16, 28 September 2008 (UTC)[reply]

While your desire to help is admirable, we don't usually do people's homework for them - they don't learn anything that way. --Tango (talk) 01:23, 28 September 2008 (UTC)[reply]

You can't look at Wikipedia articles when you're taking an exam...

I learned how to solve this problem in Calculus I. You're supposed to find a formula for the volume of the box, differentiate it, and solve it for zero within the appropriate range. (And, of course, show that that's a maximum and not a minimum!) Same thing for finding the minimal surface area of a cylinder with a given volume. You can do it other ways, but that way is more general and actually teaches you something useful. For this specific problem, let x be the length of the square you cut out from each corner, and find formulas for resulting length, width, and height of the box as a function of x. (This is basically the same answer that User:Michael Hardy gave, but it's important to think of the volume as a function.) « Aaron Rotenberg « Talk « 02:55, 28 September 2008 (UTC)[reply]

Amazing that calculus would be where anyone first learned to solve a problem of this kind. If you don't know how to solve this routinely long before you get to calculus, you're really really weak in prerequisite material. Typical, I suppose, but the blind leading the blind is what seems to happen often on this page. The stuff about differentiating is nonsense. That's what you'd do if you wanted the maximum possible volume subject to the information given. Obviously calculus is not needed to answer the question asked, which didn't say anything about finding the largest possible volume. Michael Hardy (talk) 03:16, 28 September 2008 (UTC)[reply]

I agree that the first post after yours was unhelpful, but I don't think I understand your objection to Aaron's. The OP asks for "the maximum volume of a box [subject to certain conditions]." That means the volume of the biggest possible box, subject to those conditions. I don't know what definition of maximum you're using that would allow you to stop with a formula for the volume of the box in terms of the amount cut out of the corners. Black Carrot (talk) 04:56, 28 September 2008 (UTC)[reply]

Oh---yes I did miss the word "maximum". So x(7 − 2x)(9 − 2x) is the volume when the side of the square has length x; then you need to differentiate to get the maximum of x between 0 and 7/2. 75.72.179.139 (talk) 19:07, 28 September 2008 (UTC)[reply]

(ec) The OP asked how to find the "maximum volume", so a response using the calculus is not out of place. The inequality of arithmetic and geometric means gives an upper bound which is close to the actual optimum value. However, I don't see how to get the maximum volume without using calculus. Michael Slone (talk) 05:01, 28 September 2008 (UTC)[reply]

Intemperate statements about the blind leading the blind and being really weak are not appropriate here, particularly from someone who seems to have misread the original question. To give a bit of generality on the problem, if the original paper is square, then the squares cut from each corner have one-sixth the side for maximal volume of the open-topped box. It may then be seen that the area of the base is equal to the combined area of the four sides (each 4/9 of the original area, with 1/9 discarded). And if the original rectangle has unequal sides, this property still applies, in that the expression for the base area minus the sides area is the same as the derivative of the volume function wrt x, the side of the square removed at each corner. Returning to the numerical problem given, I calculate that x = (16-root67)/6 ≈ 1.30 for greatest volume.86.152.77.53 (talk) 11:12, 28 September 2008 (UTC)[reply]

Interesting point: you don't actually need calculus for this. The local maxima and minima of polynomials can be determined completely algebraically, in most cases. Say you want the local max of x(x-5)(x-10) between 0 and 5. Simple inequalities can show that the function is positive everywhere except the endpoints, so the solution is on the interior of the interval. Subtract an arbitrary contant M, to get x^3-15x^2+50x-M, and assume M to be the maximum value of the function on the interval. This polynomial is zero at that point then, and again by simple inequalities, there must be a multiple root at that point. A fundamental fact from algebra is that if you multiply the terms of a polynomial by an arithmetic progression any multiple root of the original polynomial is still a root of the new one, and any single roots are lost. The canonical progression is the same one used in differentiation, giving 3x^3-30x^2+50x, and since 0 is not a maximum the x can be divided out. From there it's the same as with calculus. Find some roots of the polynomial and check them. Black Carrot (talk) 08:39, 29 September 2008 (UTC)[reply]

Using the Fundamental Theorem of Calculus

I already know that if you have a function

$F(x)=\int _{a(x)}^{b(x)}f(t)dt$

then $F'(x)=f'(b(x))b'(x)-f'(a(x))a'(x)$ . My question is what to do if I have

$G(x)=\int _{a(x)}^{b(x)}g(x,t)dt$ ?

Is $G'(x)=\int _{a(x)}^{b(x)}{\frac {\partial }{\partial x}}g(x,t)dt+g'(b(x))b'(x)-g'(a(x))a'(x)$ ?

What to do when the integrand is a function of t AND x? What is the most general form of the Fundamental Theorem of Calculus, the most general way of differentiating under the integral sign (or the other way around)? Thanks!--A Real Kaiser...NOT! (talk) 07:11, 28 September 2008 (UTC)[reply]

Looks good to me (assuming differentiability and all that). Anyone else? Eric. 131.215.159.210 (talk) 18:57, 28 September 2008 (UTC)[reply]

I’m not convinced – I don’t see where the other two terms are coming from. What do you mean by

g'(c)

?,

g

is a function of two variables. Also, consider this example:

G(x,t)=2xt,a(x)=0,b(x)=x

. Then

{\frac {\partial }{\partial x}}\int _{0}^{x}2xtdt=3x^{2}=_{?}x^{2}+??=\int _{0}^{x}2tdt+??

I do know if the endpoints are constants, then using the definition of an integral using a limit of a finite sum you should be able to interchange the derivative and the integral assuming differentiation is a continuous operator. (which I think it is). GromXXVII (talk) 20:23, 28 September 2008 (UTC)[reply]

The poster may possibly find Leibniz integral rule of interest. The last two terms should read g() instead of g'(). Pallida Mors 00:19, 29 September 2008 (UTC)[reply]

They still don't make sense -

g(\cdot ,\cdot )

is a function of two variables and only one has been given. --Tango (talk) 00:45, 29 September 2008 (UTC)[reply]

The full formula should read

\int _{a(x)}^{b(x)}{\frac {\partial }{\partial x}}g(x,t)dt+g(x,b(x))b'(x)-g(x,a(x))a'(x)

. Pallida Mors 03:21, 29 September 2008 (UTC)[reply]

Oh, right. Since s/he had got the F example right, I only looked at the first term. Uh... silly me. Thanks for sorting it out regardless. Eric. 131.215.159.117 (talk) 05:57, 30 September 2008 (UTC)[reply]

Thanks everyone, this is exactly what I wanted to know. I made some stupid mistakes while typing it up. Of course, G is a function of two variables. The endpoints are functions of x. And the last two terms must indeed be g instead of g'.--A Real Kaiser...NOT! (talk) 07:13, 29 September 2008 (UTC)[reply]

Differentiation of sin(ax - b) with respect to x.

The problem I am having with this is seeing two different methods, with two different answers, without seeing a flaw in any of the methods. A little guidence as to what would be the correct method of solving would be appreciated.

"sin(ax - b)" is solved using the chain rule and one ends up with "a * cos(ax - b)"

However, if one rewrites it to "-1 * sin(b - ax)" it is solved with the chain rule to become "a * cos(b - ax)"

Which one is wrong, and why? Thanks 194.81.255.156 (talk) 15:39, 28 September 2008 (UTC)[reply]

They're equal - cosine is an even function meaning cos(x)=cos(-x). --Tango (talk) 15:46, 28 September 2008 (UTC)[reply]

Thanks! Much appreciated. 194.81.255.156 (talk) 15:52, 28 September 2008 (UTC)[reply]

Thank you for letting us know that you saw and appreciated Tango's response :) mostly we never know :( -hydnjo talk 01:19, 29 September 2008 (UTC)[reply]