Answer this if your up for it!!!!!!

[Alkaline Sugar]

By cutting away identical squares from each corner of a rectangular piece of cardboard and folding up the resulting flaps, an open box may be made. If the cardboard is 15 in. long and 8 in. wide, find the dimensions of the box that will yield the maximum volume. Make sure to include the following: i) a detailed sketch of the geometry with variables; ii) define your variables; iii) formulate the quantity that you need to maximum; iv) close the interval; v) maximum the appropriate quantity using Calculus. Clearly show your detailed solution and clearly indicate your final answer. ????????????

 

[PottedMeat]

Shoulnd't you be doing your own homework?

Besides theres this new fangled thing called a 'Search Engine', in particular one called 'Google' which may be of help. In fact searching for some of the terms in the question yields an answer like this:

http://mathforum.org/library/drmath/sets/select/dm_box_maxvol.html

 

[Alkaline Sugar]

its a old home work question, bonus question of sorts. I wanted to see how to obtain the answer, thought maybe someone could tackle it.

 

[TuxDave]

I'll do the last part. x=5/3. You can do parts 1 to 5.

 

[Fallen Kell]

lol at TuxDave....

It is fairly simple when you think about it. First what is the formula for the volume of the resulting 3D object?
H*W*L = volume

So You are starting with this, then the formula using your constraints will be:

(X)*(8-(2*X))*(15-(2*X)) = Y
8>2X>0

Take the first derivative of the 1st equation above and set it equal to 0 to find the points where the slope of the tangent line is zero.

Solve the equation for X. Now using the values you obtained for X from the above

Take the second derivative of the equation and set it to zero to find critical value of the function

Using this you can prove if a point is a max or min.

Edit added ">0" in the "8>2x>0" as I forgot to put that in as it was assumed since we are dealing with real values, a negative value would mean we are somehow adding to the piece of cardboard

 

[Fallen Kell]

If you still havn't figured it out....

http://wims.unice.fr/wims/wims.cgi?sess...eft=&pright=&num_precision=12&format=t

 

[Net]

This isn't too bad.

By cutting away identical squares from each corner of a rectangular piece of cardboard and folding up the resulting flaps, an open box may be made. If the cardboard is 15 in. long and 8 in. wide, find the dimensions of the box that will yield the maximum volume. Make sure to include the following: i) a detailed sketch of the geometry with variables; ii) define your variables; iii) formulate the quantity that you need to maximum; iv) close the interval; v) maximum the appropriate quantity using Calculus. Clearly show your detailed solution and clearly indicate your final answer. ????????????

Draw a rectangle and put x's in the corners. so the length is 15-2(x) because there are two x's, etc...

1. Set up your volume function

V(x) = x(15-2x)(8-2x)


2. Multiply through and Take the first derivative
V'(X)=.......

3. use the quadratic formula to find zero's this is where is it will be maximum.

4. Pick the zero that is in your domian. Your domain is found by asking your self what numbers will work with my V(X) function. Hint we know 8 won't work because (8-2(8)) will cause the function to be negative. Why won't 0 work? Find your domain now.

5. Now use the second derivative test to see if that zero is indeed causing the function to be maximized. Is this a local maximum, global maximum or both? and why?


I worked the problem and have the answer, post your results here and I'll tell you if you got it. Good luck.

 

[Fallen Kell]

Originally posted by: net
This isn't too bad.

By cutting away identical squares from each corner of a rectangular piece of cardboard and folding up the resulting flaps, an open box may be made. If the cardboard is 15 in. long and 8 in. wide, find the dimensions of the box that will yield the maximum volume. Make sure to include the following: i) a detailed sketch of the geometry with variables; ii) define your variables; iii) formulate the quantity that you need to maximum; iv) close the interval; v) maximum the appropriate quantity using Calculus. Clearly show your detailed solution and clearly indicate your final answer. ????????????

Draw a rectangle and put x's in the corners. so the length is 15-2(x) because there are two x's, etc...

1. Set up your volume function

V(x) = x(15-2x)(8-2x)


2. Multiply through and Take the first derivative
V'(X)=.......

3. use the quadratic formula to find zero's this is where is it will be maximum.

4. Pick the zero that is in your domian. Your domain is found by asking your self what numbers will work with my V(X) function. Hint we know 8 won't work because (8-2(8)) will cause the function to be negative. Why won't 0 work? Find your domain now.

5. Now use the second derivative test to see if that zero is indeed causing the function to be maximized. Is this a local maximum, global maximum or both? and why?


I worked the problem and have the answer, post your results here and I'll tell you if you got it. Good luck.

hehehe... someone didn't read my last 2 posts....