

Subject:
CalculusOptimization problem
Category: Reference, Education and News > Homework Help Asked by: jktallnycga List Price: $2.50 
Posted:
08 Dec 2005 15:28 PST
Expires: 11 Dec 2005 14:19 PST Question ID: 603399 
An optimization problem: If 1200 centimeters squared of material is available to make a box with a square base and an open top, find the largest possible volume of the box. 

There is no answer at this time. 

Subject:
Re: CalculusOptimization problem
From: jonshaga on 08 Dec 2005 19:12 PST 
Here's how to do the problem: Let the length, width, and height of the box (in cm) be represented by x, x, and y (because the base is a square, the length and width must be equal): V = x^2*y (volume formula) We can express V as a function of just one variable x by using the fact that the area of the four sides and the bottom of the box is: 4xy + x^2 = 1200 Solving for y, we get: (1200  x^2)/4x = y Now plug in to the volume formula: V = x^2*y = x^2*(1200  x^2)/4x = 300x  (x^3)/4 Now comes the main part of the problem. As you may or may not recall, finding maximum/minimum values ("extremes") requires the use of partial derivatives (because we are only working with one variable, this is more of a regular derivative): dV/dx = 300x  (x^3)/4 0 = 300x  (x^3)/4 0 = x(300  (x^3)/4) It's 0 only when x = 0 or x = 10.62 Because a side length can't be zero, x (length & width) must be 10.62cm. So, the box will have a maximum volume when both the length and width are 10.62 cm. The height (y) can be found using the formula derived above: y = (1200  x^2)/4x = 25.59cm Now that we know the dimensions, we can find the volume: V = x^2 * y = 2886.2 cm^3 Hope this helps! (and you didn't have to pay me $2.50) 
Subject:
Re: CalculusOptimization problem
From: myoaringa on 08 Dec 2005 20:21 PST 
The largest volume box will be a cube, i.e., bottom and four sides of equal size and square. 1200/5 = 240 Square root of 240 = 15.49, the length of the sides, ^3 = 3718 cubic cm. Of course, you will have to cut and paste a bit to use all the material. How does one know that a cube is the optimal shape? A sphere is the figure with the best volume to surface ratio, so logic (common sense) indicates that the rectilinear figure closest to a sphere in proportions will have the largest volume to surface ratio. Or can you prove me wrong? 
Subject:
Re: CalculusOptimization problem
From: jktallnycga on 08 Dec 2005 21:18 PST 
jonsha Thanks, your answer helped me out of my mental block! Even though when you took the derivative of the voume equation, it should be dV/dt = 300  3/4 x^2 and not what you put, dV/dx = 300x  (x^3)/4 But, I was able to figure out myself where it went wrong. As for myoarinI'm not sure you ever have done one of these problems. At least not for a Calculus class. Maybe your common sense told you "that the rectilinear figure closest to a sphere in proportions will have the largest volume to surface ratio," but my common sense told me to use a box because the words in the problem clearly ask to solve the problem using a figure that is a box. 
Subject:
Re: CalculusOptimization problem
From: ansel001ga on 08 Dec 2005 23:41 PST 
Here is the answer. Jonshua started out correctly. Let the length, width, and height of the box (in cm) be represented by x, x, and y (because the base is a square, the length and width must be equal): V = x^2*y (volume formula) We can express V as a function of just one variable x by using the fact that the area of the four sides and the bottom of the box is: 4xy + x^2 = 1200 Solving for y, we get: (1200  x^2)/4x = y Now plug in to the volume formula: V = x^2*y = x^2*(1200  x^2)/4x = 300x  (x^3)/4 Now comes the main part of the problem. As you may or may not recall, finding maximum/minimum values ("extremes") requires the use of partial derivatives (because we are only working with one variable, this is more of a regular derivative): Here is where a correction needs to be made V = 300x  (x^3) / 4 dV/dx = 300  3/4 *x^2 = 0 3/4 *x^2 = 300 x^2 = (4/3)* 300 = 400 x = 20 y = (1200  x^2)/4x = (1200  (20^2) / (4 * 20) = (1200  400) / 80 = 800 / 80 = 10 V = x^2*y = (20^2) * 10 = 400 * 10 = 4000 As a check on the answer Surface area of the box 4xy + x^2 = (4*20*10) + (20^2) = 800 + 400 = 1200 Hope this helps. 
Subject:
Re: CalculusOptimization problem
From: myoaringa on 09 Dec 2005 08:15 PST 
OH well, so much for my kind of common sense! :/ 
Subject:
Re: CalculusOptimization problem
From: ansel001ga on 09 Dec 2005 17:46 PST 
Myoarin, For a closed box, it would have been in the shape of a cube, but there was no top. Think of it this way  you want to maximize the volume you get for a given surface area. You have walls and bases (top and bottom). You have to enclose all the walls (y variable) but only one of the bases (x variable), namely the bottom. Consequently you get more bang for your buck (volume for your surface area) when you increase the base than you do when you make the walls taller. 
Subject:
Re: CalculusOptimization problem
From: myoaringa on 10 Dec 2005 04:28 PST 
HI, Anse1001, Thank you for explaining the error in my thinking. Of course ...! Myoarin 
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