Largest Possible Volume of a Box Problem.?

Question

If 1500 square centimeters of material is available to make a box with a square base and an open top, find the largest possible volume of the box.


Volume = _____________ cubic centimeters.


Can someone help me solve this problem?

Answer 1

Let the side of a base be x.
Let the height be h.

The Surface area is 1500. so x^2 + 4xh = 1500
and h = (1500 - x^2)/4x.

Volume of the box is (x)(x)((1500 - x^2)/4x)

Simplifying yields the function y = (1500x - x^3)/4

I cheated and used my graphing calculator to determine the maximium. It occurs at 5590.1699

Answer 2

If the box has a square base, then the volume is x*x*h, where x is the edge length of the square base and h is the height. The total surface area of the box is:
bottom + four sides = x^2 + 4(xh)

This has to be 1500, so if we pick a value for x, that forces us to pick a value for h, and vice versa. So we can write h in terms of x by using the expression for the total surface area:
x^2 + 4(xh) = 1500
4(xh) = 1500 - x^2
h = (1500 - x^2) / 4x

So the volume of the box is always going to be:
V = x*x*h
V = (x^2) * ((1500 - x^2) / 4x)
V = (x) * ((1500 - x^2) / 4)
V = 375x - (1/4)x^3

If we were to graph this, we'd see some parts where the curve reaches a peak and then comes back down. We want to find the values of x that give us a peak, because that means the volume is maximized. At the of a rolling peak on a graph, the slope of the tangent is zero. So we need to take the derivative of V and set it equal to zero, to find the x that gives us the biggest volume.

V'(x) = 375 - (3/4)x^2
375 - (3/4)x^2 = 0
(3/4)x^2 = 375
x^2 = 500
x = sqrt(500)

Remember we found that V = 375x - (1/4)x^3, so plug in this value of x to find the volume.

It's worth noting that the solution is NOT a cube. That's not the shape that gives a maximum volume to area ratio when we're considering are of materials of an OPEN box.

Answer 3

Let the square base have side s, and let the height of the box be h. Then the surface area of the box will be

s^2 + 4sh = 1500 (square base + 4 rectangular sides)

since the box has an open top.

This constraint must be satisfied for any s and h which are solutions. The problem then is to maximize the volume V = s^2h, given this constraint. Through some simple manipulation of the surface area constraint we get

h = (1500 - s^2)/4s

so V = s^2(1500-s^2)/4s = (1500s -s^3)/4

Extrema will be at dV/ds = 0, so

dV/ds = (1500 -3s^2)/4

This is 0 for s^2 = 500, i.e. s = 10sqrt(5)

d/ds(dV/ds) = -6s which is negative at this value, so this is a maximum of the volume function (negative second derivative)

If s = 10sqrt(5) then h = 1000/(40sqrt(5)), plugging into the formula for h. Simplifying

h = 25/sqrt(5) for the s that maximizes V.

So the maximum volume V = s^2h, plugging the values for s and h we just found, is

V = 12500/sqrt(5)

Answer 4

This is an optimization problem. The tricky part is to translate it into the right set-up. (It turns out that we don't need partials btw --- the constraints given make it a 1-dimensional problem) We want to figure out the shape of the box --- its width w, length l, and height h. How much volume a box with width w, length l, and height h contains, is: V(w, l, h) = w*l*h And we have the constraint that the material we use is 200 inches^2: 2*w*h + 2*l*h + 2*w*l + w*l = 200 (*) We added the extra w*l because the bottom of the box is reinforced. We also have the constraint that the box has to have square ends i.e., that the length l = height h. So if we know the length l of a box, then we know its height; from this and (*) we can figure out its width. Thus the shape of the box depends only on its length. From the length we figure out: w = l h = (200 - (3*w*l))/(2*w+2*l) = (200 - (3*l*l))/(4*l) So the volume of a box with length l is: V(l) = l*l*(200 - (3*l*l))/(4*l) = -3/4*l^3 + 50*l. Set the derivative equal to 0, and solve. Hope this helps.

Answer 5

Let the base have side length x cm and the height be h cm. The area required is x^2 + 4xh = 1500, so h = (1500-x^2)/4x. The volume is V = x^2.h = 375x - x^3/4.
So dV/dx = 375 - (3/4)x^2 and d2V/dx2 = -3x/2.
dV/dx = 0 <=> x^2 = 4(375)/3 = 500
<=> x = ±10√5. But we need x > 0, so x = 10√5.
At x = 10√5, d2V/dx2 = -15√5 < 0, so this represents a maximum as expected.
V = 375x - x^3/4
= 375(10√5) - (10√5)^3 / 4
= 3750√5 - 1250√5
= 2500√5 cm^3
≈ 5590 cm^3.

Answer 6

square base, call x and call height y
Surface Area = x^2 + 4xy =1500
solve for y ===> y= (1500 - x^2) / (4x)
Volume V = x^2 y
sub in y ===> V = x^2 (1500 - x^2) /(4x)
simplify ===> V = 375 x -(1/4)x^3
now set V ' = 0 to get max

V ' = 375 -(3/4) x^2 = 0 ===> x^2 = 500 ===>x>0 ===> x=sqrt500, which reduces to 10sqrt5

x=10sqrt5, solve for y ===> y= (1500 - x^2) / (4x)
y = 1000 / (40sqrt5) = 5sqrt5

Volume = 10sqrt5 * 10sqrt5 * 5sqrt5 = 2500sqrt5

Answer 7

I believe the largest volume is a cube .
surface area is 5 x S * S where S is the length of one side
( 5 x because of open top )
1500 = 5 x S^2
300 = S^2
S = 17.32

volume = S x S x S
= 5196 cubic centimeters

Answer 8

Let sides of the top and bottom be x
Let the height of box be h

Then for the material,
x^2 + 4xh = 1500 (x^2 are for the top, xh for the sides)

Also, volume, v= hx^2

From the 1st equation, rewrite for h
h = (1500 - x^2)/(4x) = (375/x) - (x/4)

so now v = (x^2)[(375/x) - (x/4)]
v = 375x - [(x^3)/4]

For max, dv/dx = 0
so,
dv/dx = 375 - 3(x^2)/4 = 0
solving,
x = 22.36 giving h = 11.18

Hence, volume = 5589.66 cm^3

Answer 9

i'm trying to remember why but i know, a cube makes the largest volume. so if it has an open top, just divide the total area by 5.
1500/5=300. each face has to have an area of 300. so the length of each side has to be sqrt(300). so the volume is the length of side cubed. V=sqrt(300)^3

Answer 10

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