An open box is to be made from a 3-foot by 5-foot rectangular piece of material by cutting euqal squares from?

Question

An open box is to be made from a 3-foot by 5-foot rectangular piece of material by cutting euqal squares from each corner and turning up the sides. Find the volume of the largest box that can be made in this manner.

(a). 5.2ft³ (b). 4.1ft³ (c). 7.5ft³ (d). 3.3ft³

PS: can any1 show me how to do this problem step-by-step...thanks

Answer 1

Let side of each square = x ft
Box then has dimensions volume (ft³):-
V(x) = (5 - 2x)(3 - 2x)(x)
V(x) = (15 - 16x + 4x²) (x)
V(x) = 15x - 16x² + 4x³
V `(x) = 15 - 32x + 12x² = 0 for max V
x = 2.06 , x = 0.61
V(0.61) = 3.78 x 1.78 x 0.61 = 4.1

OPTION b

Answer 2

Z = how much you cut from each side. ..... the amount you cut is going to end up being teh height.

V = L * W * Z
L = 3-2Z
W= 5-2Z
Ok so lets say you start on the 5 foot side.
From the left you cut in at 1 foot........thats Z......you're going to have to cut one inches from the right side also. You got a nice little flap. Well to make the edges meet.....on the 3 foot side you're also goign to have to cut in 1 foot on each side.
Just try to visualize that.


(3-2Z)(5-2Z)(Z) = V
15Z--16Z^2 + 4Z^3 = V
take the derivative
12 Z^2 -32Z +15 = 0

Solve for Z. using quadratic equation..... you get 2 answer.....

Z = .61
and Z = 2..06.
Z can't be 2.06......the 3 foot side of your paper....if you cut 2 foot from one side and 2 from the other.....well thats impossible.

Now plug Z =.61 back into your equation
(3-2Z)(5-2Z) * Z = 4.1 ft^3

Answer 3

Let's cut a square of size x by x from each corner.

This means the base will 5-2x by 3-2x, and the height will be x

Volume = f(x) = (5-2x)(3-2x)(x)
= (15 - 16x + 4x^2)x
= 15x - 16x*2 + 4x^3

Max volume will occur when f'(x) = 0

f'(x) = 15 - 32x + 12x^2 = 0
x = (32 +/- sqrt(32^2 - 4(12)(15)) / 2(12)
= (32 +/- sqrt(1024 - 720)) / 24
= (32 +/- sqrt(304)) / 24
= (32 +/- 4 sqrt(19))/ 24
= (8 +/- sqrt(19))/6
= 2.06 or 0.61

Now we know the base is 5-2x by 3-2x, so we want x = 0.61

Volume = (5-2*.061)(3-2*0.61)(0.61) = 4.10

Answer is b

Answer 4

If you cut a square from each corner. Therefore let each side of each square equal x. therefore
Area box = (5-2x)(3-2x)x
= (15-6x-10x+4x^2)x
= 4x^3-16x^2+15x
Therefore dA/dx= 12x^2-32x+15
To find the maxima dA/dx=0
0=12x^2-32x+15
x=2.06,0.61

x 2.05 2.06 2.07
dA/dx - 0 +

This proves that 2.06 is a minimum, therefore to find the maximum area of the box, x must be equal to 0.61

Therefore we put this back into the original area equation:
Area box = (5-2x)(3-2x)x
=4.1 ft cubed

Answer 5

V = L * W * H

H = side of the square to be cut out
W = 3 - 2H (since you cut on both sides)
L = 5 - 2H (since you cut on both sides

V = (5-2h)(3-2h)(h)
V = 15h -16h² + 4h³
dV/dH = 15 - 32h +12h² = 0
12h² - 32h + 15 = 0
using quadratic formula:
h = 0.60685 ; h = 2.0598 (not valid coz we can't cut 2x2.0598 on the 3ft width


V @ h= 0.60685
V = [5-2(0.60685)] [3-2(0.60685)] [0.60685]
V= 4.104410368 ft³

so the answer is b