optimization problem for calculus?

Question

i hate these problems and never have the patience or knowhow to figure them out, but i think this one is relatively simple for anyonewho knows what thyre doing.

A box with a square bottom and open top is to be made from 27 square feet of material, what is the largest possible volume of the box.

I could really use help on this, its an example of a question that will be just like it on my final tomorrow so if someone could help id be very greatful.

Answer 1

First off, you're given the surface area of the box with a square bottom and open top. The formula for the surface area with a square base and open top is below:

S = (area of base) + (area of sides)
Let x = length of box. [note: x is also the width of the box since the base is square)
y = height of box.

Then, since there's one square base and 4 sides, we have

S = x^2 + xh + xh + xh + xh
S = x^2 + 4xh

However, we're given S = 27; therefore

27 = x^2 + 4xh

What we're trying to do is maximize the volume. The volume of this particular box is given by the formula:

V = (x^2)h

But, we can calculate h in terms of x, from here:

27 = x^2 + 4xh

27 - x^2 = 4xh
(27 - x^2)/(4x) = h

Therefore, we replace the variable h in the formula for the volume.

V = (x^2) [ (27 - x^2)/(4x) ]

Which we can reduce to

V = x (27 - x^2)/4
V = (27x - x^3)/4

Now, we can declare this our volume function V(x).

V(x) = (27x - x^3)/4

In order to find the maximum volume, we have to take the derivative V'(x) and make it 0.

V'(x) = (1/4) [27 - 3x^2]

0 = (1/4) [27 - 3x^2]
0 = 27 - 3x^2

Moving the -3x^2 over to the left hand side,

3x^2 = 27
x^2 = 9
x = +/- 3

Note that we have two solutions: x = -3 and x = 3. However, we can discard x = -3 since we can't have a side of negative length.

To find the largest possible volume of the box, all we have to do is solve V(3).

V(3) = ( 27(3) - 3^3 ) / 4 = ( 27(3) - 27 ) / 4
V(3) = 54/4 = 27/2 = 13.5

Answer 2

You need a variable first. Obvious choice here seems the side of the square. Call this x.

Work with the surface area first suppose the height is h then the surface area is aa for the bottom and 4ah for the sides (4 sides to the box). This is the amount of material you can use and can be used to work out h

a^2 + 4ah = 27

4ah = 27 - a^2

h = (27 -a^2)/4a

The volume of the box is area of base times height, i.e.

a^2 * (27-a^2)/4a =a (27-a^2)/4 = (27a-a^3)/4

Now you have a formula for the volume of the box you need to know how to maximise this by varying a, so use differential calculus, i.e. set the derivative to zero.

f'(a) = (27-3a^2)/4 = 0
multiply each side by 4

27 - 3a^2 = 0

rearrange
27 = 3a^2
9=a^2

so a=3 or -3 are possible solutions.

Plug this into the volume formula above to get the volume

(27*3-3^3)/4 = 13.5

The -3 gives an answer opposite in sign to this.

To really show off you could take the second derivative of the box volume to show it is a maximum.

Answer 3

Let x = width of the box = length of the box.
Let y = height of the box.

The (outer) surface area of the box is x^2 + 4xy,
which should be 27, so we have the constraint condition
x^2 + 4xy = 27.

The volume of the box is v(x, y) = yx^2.

I am assuming that this problem is from lower-level calculus,
so I wouldn't use Lagrange multiplier method.

We manipulate the constraint condition to obtain
4xy = 27 - x^2
-> y = (27 - x^2)/4x if x is not 0 (which we can assume without
loss of generality since a box with the width 0 is physically
impossible).

Now we plug y = (27 - x^2)/4x into the volume function to obtain

v(x, y) = [(27 - x^2)/4x]x^2 = x(27 - x^2)/4, so now we
have made the volumn function into a function with only
one independent variable. Let us rewrite

v(x) = (27/4)x - (1/4)x^3.

To maximize or minimize v(x), we differentiate to find the
critical point of v(x). The derivative of v(x) is

v'(x) = 27/4 - (3/4)x^2

which is 0 if and only if x^2 = 9. Hence, v(x) is maximized
when x = 3 (we obviously drop the case x = -3).
By plugging into the constraint, we find that y = 3/2.

Thus if the dimension of the box is 3 x 3 x 3/2, then the
volume is maximized, and its volume would be 3 * 3 * 3/2
= 27/2.

I haven't checked the work, so I will leave it to you to check
computation.

Answer 4

Sorry, i think misprints (must be "inscribed" instead of "circumscribed" in accordance the parent and "maximize" instead of "cut back") contained in the project, making it mindless. around a given sphere you may circumscribe a cone with volume, which would be as super as you opt for, think of its top increasing infinitely making its vertex further and extra sharp going up, or, although, think of cone radius to improve infinitely, reducing the top to suggestions-set the sector diameter. the two approaches the quantity of the cone will improve infinitely. The yet another situation (to cut back the quantity of the circumscribed cone) is incredibly exciting certainly and the respond is r/h = a million/sqrt(8) = a million/(2sqrt(2)). that's precisely the cotangent of the angles (say ?, 0 < ? < ?/2) to the backside of the isosceles triangle on the parent, then cos (?) = a million/3, which skill ? equals the dihedral perspective of the widely used tetrahedron, so the optimal answer could be defined for that reason: the cone with the least volume is inscribed in a widely used tetrahedron, circumscribed appropriate to the given sphere. to get carry of that take volume(cone)/volume(sphere) = (a million/3)*?*r^2*h/((4/3)*?*a^3) = (a million/4)*(r/a)^2*(h/a) = (a million/4)*(r/a)^3*(h/r) = (a million/4)*cot^2(?/2)*tan(?) = a million/(2*tan^2(?/2)*(a million - tan^2(?/2))) Taking x = tan^2(?/2) /0 < x

Answer 5

you need two equations I would suggest start off with area

area is XY=27 the other one you need is perimeter so
p= 2x+2y

this should give you an idea of some sort