A square sheet of cardboard 45 inches on a side is made into a box by cutting squares of equal size from each corner of the sheet and folding the projecting flaps into an open-top box. What should be the length of the edge of any of the cutout squares to give the box maximum volume?
The book list many answers. I have tried this several times and come up with 5.5 inches. What do you get when you solve?
a. 4 inches
b. 4.5 inches
c. 5.5 inches
d. 6.5 inches
e. 7 inches
f. 7.5 inches
g. 8 inches
h. none of these
2006-12-10 03:44:37 · 6 answers · asked by chris 2 in Science & Mathematics ➔ Mathematics
The shape that maximizes volume is a square, which means that the side should be 1/3 of 45 (15) so that two sides can be folded up.
Answer is H. none of these.
2006-12-10 04:03:10 · answer #1 · answered by The J Man 2 · 0⤊ 1⤋
Call the edge length x. V = LWH. L and W = 45 - 2x; H = x. Therefore V = x(45-2x)^2 = x(2025 -180x + 4x^2) = 2025x -180x^2 + 4x^3.
We want to maximize V, so first we have to take the derivative of it. V' = 12x^2 - 360x + 2025. Now, we set this equal to zero, making it a quadratic equation. x = (360 +/- sqrt(360^2 - 4*12*2025))/2*12. Solve the square root first: x = (360 +/- sqrt (32400))/2*12 = (360 +/- 180)/24. That means x = 180/12 or 540/12 = 15 or 45. Obviously 45 can't be right, so it must be 15. That makes sense to me, as it makes the box a cube, which is generally ideal for maximizing volume.
The length is 15 inches.
2006-12-10 03:58:28 · answer #2 · answered by Amy F 5 · 0⤊ 1⤋
You can solve this as a "thought experiment" rather than just using math; Think about what shapes 'maximize volume', and you'll realize the most effective shape to do that is a sphere. Since we have to make an ope-topped box here, the equivalent would be an open-topped cardboard cube. Therefore the answer will be which ever one of these options gives you a box closest in shape to a cube.
I'll let you do the rest!
2006-12-10 03:50:22 · answer #3 · answered by Anonymous · 0⤊ 2⤋
I got an answer of 15in for each side this is what i did.
first you find that the lenght of each side is going to be (45-x)
x is the part left after you cut it.
second, since you know that the volume of a cube is (lenght *width*height) you can use (45-x) ^3.
third, you derive that equation and then set it equal to zero.
Your answer for that is going to be (45) the maximum volume.
Then if you divide that by three you get the maximum lenght of each side.
2006-12-10 04:07:01 · answer #4 · answered by Felix_Da_Cat 2 · 0⤊ 0⤋
Amy F her approach is correct. However, it is 180/24 and 540/24 not 12. The 15x15x15 dimension gives u 3375 where 30 x 30 x 7.5 gives you 6750
2006-12-11 03:36:32 · answer #5 · answered by eretzkid85 1 · 0⤊ 0⤋
Draw the image! The ladder varieties a dazzling triangle with the wall and the floor. The hypotenuse is a continuing=29 ft. The apex of the dazzling attitude is on the beginning place. the top =y So, by using the Pythagorean Thm., the backside is x = (29^2 - y^2)^0.5 we are seeking the fee of displacement of x , or dx/dt So, because of the fact that all of us understand x in terms of y, we are in a position to apply that expression to discover the 1st spinoff by using the chain rule:: dx/dt= d[(29^2 - y^2)^0.5] /dt dx/dt= [0.5 * (29^2 - y^2) ^(-0.5)] * (-2y) * dy/dt we are provided that dy/dt = -2 ft consistent with 2d and y= 20 ft, so replace those values, and plug and chug: hence, dx/dt= 40/21 ft consistent with 2d while y=20.
2016-10-18 01:44:45 · answer #6 · answered by Anonymous · 0⤊ 0⤋