An open box is to be made from a 16 inch by 30 inch piece of cardboard?

Question

An open box is to be made from a 16 inch by 30 inch piece of cardboard by cutting equal squares from each corner and folding up the sides. what size should the squares be in order to obatin a box with the largest possible volume?

Answer 1

assume the squares you cut off at the corners have sides of "x" length

the volume of your box (V) = (30-2x)*(16-2x)*(x) = 4x^3 -92x^2 +480x

when you graph this equation, you will get the maximum and minimum volumes.

Without graphing: remember that maximum and minimum values happen when the graph is changing direction? i.e. the slope is equal to zero. Knowing that, we can take the derivative of the equation: dv/dx = 12x^2 - 184x + 480 = 0
--> 3x^2 - 46x + 120 = 0
--> x^2 - (46/3)x = -120/3 = -40
--> (x - 23/3)^2 = -40 + (23/3)^2 = 169 / 9
--> x - 23/3 = +/- 13 / 3
--> x = 12 (not valid) or 10/3
==> x = 10/3 inch

Answer 2

comic strip out this concern (or maybe decrease out the corners of a sheet of cardboard to persuade your self of this)... in case you decrease out an x by using x sq. from each and each nook, and then fold up the tabs, the dimensions of the field would be: V(x) = x(sixteen - 2x)(30 - 2x) <<2x because of the fact there are squares decrease from each and each end of a edge>> word additionally the area of x: (0 , 8)... that's mandatory to make specific the two edge of the field is beneficial that's the function to maximise V(x) = (16x - 2x^2)(30 - 2x) V(x) = 480x - 32x^2 - 60x^2 + 4x^3 V(x) = 4x^3 - 92x^2 + 480x V ' (x) = 12x^2 - 184x + 480 V" (x) = 24x - 184 for 0 < x < 8, V" would be < 0, confirming that this may well be a max to discover nearby max, set V' = 0 12x^2 - 184x + 480 = 0 4(3x^2 - 46x + one hundred twenty) = 0 4(x - 12)(3x - 10) = 0 so the two x = 12 (exterior the area) or x = 10/3 = 3 one million/3 the optimal volume would be with a ten/3 = 3 one million/3 in sq. decrease out of each nook sixteen - 2(10/3) = 40 8/3 - 20/3 = 28/3 = 9 one million/3 30 - 2(10/3) = ninety/3 - 20/3 = 70/3 = 23 one million/3 dimensions of max volume: 3 one million/3 by using 9 one million/3 by using 23 one million/3 in. max volume: 725.9 in^3 (shown graphically)

Answer 3

Since our cardboard is 16 by 30, and we're cutting squares of unknown sizes (this is much easier represented by a diagram), let's denote one side of the square to be x.

Then, we envision length times width times height.

The length is going to be 30 - 2x, since two sides of a square are taken.
The width is going to be 16 - 2x, since two sides of a square is taken.
The height is going to be x, since that's how much we fold it by.

Therefore,

V(x) = (30 - 2x) (16 - 2x) (x)
V(x) = (480 - 92x + 4x^2)x
V(x) = 480x - 92x^2 + 4x^3

To maximize the volume, we take the derivative and make it 0.

V'(x) = 480 - 184x + 12x^2
0 = 480 - 184x + 12x^2

We can divide by 4

0 = 120 - 48x + 3x^2
0 = 3x^2 - 48x + 120

We can divide by 3

0 = x^2 - 16x + 40
0 = x^2 - 16x + 64 - 44
0 = (x-8)^2 - 44
44 = (x-8)^2
(x-8) = +/- sqrt(44)
(x-8) = +/- 2 sqrt(11)
x = 8 +/- 2 sqrt(11)

We discard the negative root, since x can't be negative.

Therefore, x = 8 + 2sqrt(11)

Answer 4

7by 7

i think