Help me with me Calculus Homework pls (Optimization)

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TheTuanster

Senior Member

11-19-2013

1. An open box is to be made from a rectangular piece of material by cutting equal squares from each corner and turning up the sides. Find the dimensions of the box of maximum volume if the material has dimensions 6x6 inches.

2. An open box is to be constructed from cardboard by cutting out squares of equal size in the corners and then folding up the sides. If the cardboard is 6x11 inches, determine the volume of the largest box which can be constructed.

I absolutely have no idea where to start.


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The Fatalii

Senior Member

11-19-2013

Quote:
Originally Posted by TheTuanster View Post
1. An open box is to be made from a rectangular piece of material by cutting equal squares from each corner and turning up the sides. Find the dimensions of the box of maximum volume if the material has dimensions 6x6 inches.

2. An open box is to be constructed from cardboard by cutting out squares of equal size in the corners and then folding up the sides. If the cardboard is 6x11 inches, determine the volume of the largest box which can be constructed.

I absolutely have no idea where to start.
1. Karthus

2. Im pretty sure its still Karthus (let me double check my work)

yupp Karthus


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Bad Nikes

Member

11-19-2013

Quote:
Originally Posted by TheTuanster View Post
1. An open box is to be made from a rectangular piece of material by cutting equal squares from each corner and turning up the sides. Find the dimensions of the box of maximum volume if the material has dimensions 6x6 inches.

2. An open box is to be constructed from cardboard by cutting out squares of equal size in the corners and then folding up the sides. If the cardboard is 6x11 inches, determine the volume of the largest box which can be constructed.

I absolutely have no idea where to start.
Always start this kind of problem by setting up a formula to work with.

For #1, the volume of the resulting cube is going to be (6-2x)^3 where x is the length of the cutout in inches. Work from there.

Edit: Actually that equation is completely wrong. Gimme a sec.


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TheTuanster

Senior Member

11-19-2013

Quote:
Originally Posted by The Fatalii View Post
1. Karthus

2. Im pretty sure its still Karthus (let me double check my work)

yupp Karthus

Ok thankyou! I had no idea it was that simple.


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Mafia Goat

Senior Member

11-19-2013

im so dumb i literally stared at this and in my head said "yup a box, that's a good box"


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Durp Purple

Member

11-19-2013

Quote:
Originally Posted by The Fatalii View Post
1. Karthus

2. Im pretty sure its still Karthus (let me double check my work)

yupp Karthus
Rofl


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Sandageto

Senior Member

11-19-2013

pay attention in class then retarded downie

calc was years ago for me and i can tell you it has to do with taking a derivative to find the max and min


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Crastym

Senior Member

11-19-2013

Let's see.... #2

V = (6-2z) * (11-2z) * z where z is the length of one of the sides of the square, which also happens to be the height of the box.

Take the derivative of V with respect to z, set dV equal to 0, then solve for z. You'll want to expand V before you take the derivative.


Let me know what you get?


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True Vanity

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Senior Member

11-19-2013

Quote:
Originally Posted by Bad Nikes View Post
Always start this kind of problem by setting up a formula to work with.

For #1, the volume of the resulting cube is going to be (6-2x)^3 where x is the length of the cutout in inches. Work from there.

Edit: Actually that equation is completely wrong. Gimme a sec.
V = (L)(W)(H) = (6-2x)(6-2x)(x)


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Bad Nikes

Member

11-19-2013

Ok I suck at visualizing this kind of thing so I had to draw it out. The equation for problem 1 that you will need to maximize is

V=x(6-2x)^2

Edit: shenned, but yeah I think you get the idea. In calculus you need to know how to manipulate equations obviously, but also to know how to set them up from raw data.


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