MATH help... Integral Calculus

[b fold the truth]

Ok...a buddy (aka Cheese for those who visit the Traning Board...hehe) came over for some Math help tonight. Of course...if it don't involve 10 fingers and 10 more toes...I'm lost and no help whatsoever!!!!

Was hoping that a few of the brains on this board could help with a few answers on a couple problems.

#3. 10 points
An open box is to be made by cutting equal squares for each corner of a 12-in square piece of cardboard and then folding up the sides. Find the length of the side of the square that must be cut out if the volume of the box is to be maximized.
What is the maximum volume?




#4. 8 points
The height of an object moving vertically is given by s=-16t^2 +96t+112 with "s" in feet and "t" in seconds. Find the following:
a. The time it takes the object to reach it maximum height.
b. The maximum height of the object.




#9. 10 points
Find the approximate area under the curve of y= (x)^(1/2), between x=1 and x=4, by dividing the indicated intervals into n=12 subintervals and then add up the areas of the inscribed rectangles.

I, as well as Cheese, appreciates any and all help on these problems. They are due on Tuesday morning at 8am. Thanks...!!!!!

B True

 

[prophet]

#3

12/6=2

2^3=8 cubic inches

 

[prophet]

#3

12/6=2

2^3=8 cubic inches

sorry bro i have no clue i don't think that works

 

[Norman Bates]

4. s= -16tt +96t +112
s´=-32t +96
lokales extremum: s´=0
s´= -32t +96 =0
-32t =-96
t=3
maybe you will have to prove that it is a maximum and not the minimum, don´t know for sure.

now insert t=3:

s=-16 *3*3 +96*3 +112= 256

 

[aalox]

#3

volume (V) = (12 - 2*S) * (12 - 2*S) * S, where S is the length of a side of the squares being cut out
V = 144S - 48S^2 + 4S^3
I should add that this only makes sense if S > 0 and S < 6.
to find the max, find the derivative of V with respect to S and set that to 0
dV/dS = 144 - 96S + 12S^2 = 0
12 - 8S + S^2 = 0
(S - 2)(S - 6) = 0
S could be 2 or 6, but if six-inch squares were cut out, there would be no material left! If you were to plot V over S, it is actually a local minimum.

Therefore S must be 2.

volume = (12 - 2*2) * (12 - 2*2) * 2

volume = 128 cubic inches

 

[prophet]

#3

volume (V) = (12 - 2*S) * (12 - 2*S) * S, where S is the length of a side of the squares being cut out
V = 144S - 48S^2 + 4S^3
to find the max, find the derivative of V with respect to S and set that to 0
dV/dS = 144 - 96S + 12S^2 = 0
12 - 8S + S^2 = 0
(S - 2)(S - 6) = 0
S could be 2 or 6, but if six-inch squares were cut out, there would be no material left! If you were to plot V over S, it is actually a local minimum.

Therefore S must be 2.

volume = (12 - 2*2) * (12 - 2*2) * 2

volume = 128 cubic inches

how is that possible with a 12 inch square to start with?

 

[aalox]

Norman Bates is correct on #4.

#9

Pick 12 evenly spaced spans between 1 and 4, starting with 1. They are x = 1 to 1.25, x= 1.25 to 1.5, x = 1.5 to 1.75, etc.
Evaluate the function at x = 1, 1.25, 1.5, etc.
sqrt(1) = 1, sqrt(1.25) = 1.118, sqrt(1.5) = 1.225, etc.
Multiply each result by the x dimension, 0.25.
Then add them all up.
0.25 + 0.2795 + 0.3062 + ...

That should give you your answer.

 

[aalox]

how is that possible with a 12 inch square to start with?

In my solution, the "flaps" are 8" by 2" and the "base" of the box is 8" by 8". Fold the flaps up, and the box is 2" "high."

 

[prophet]

In my solution, the "flaps" are 8" by 2" and the "base" of the box is 8" by 8". Fold the flaps up, and the box is 2" "high."

think about it.....you have 1 square piece of cardboard....its not possible to get that many cubic inches out of a piece that small....remember a cube has 6 sides

 

[aalox]

think about it.....you have 1 square piece of cardboard....its not possible to get that many cubic inches out of a piece that small....remember a cube has 6 sides

The problem calls for "an open box". I understand that to mean a box with only five sides of material; no material is used on the sixth side to seal the box.