Maxima/Minima Problem in Calculus...?

Question

This problem is bothering me!
A box is to be made from:
-a piece of cardboard that is 9 sq units in area
-by cutting away square edges
Q: Find the maximum volume that can be made out of this cardboard.

*Seems like an easy problem but i have no idea how to relate the length of the square with the sides of the cardboard. Please help! Thanks!

Answer 1

Assuming the piece of cardboard is a square, each side is 3 units long:

If the side of the cut-away squares are x units long, the area of the base of the box will equal : (3-2x)(3-2x)

To get the volume, you multiply this with the height of the box (x) and get:
(3-2x)(3-2x) * x = 9x - 12x^2 + 4x^3

To find the max you differentiate, and get:
dy/dx = 9 - 24x + 12x^2 = 0 (Edit: I had done this on paper and copied it wrong)

You know that x has to be a number between 0 and 1.5 (2x < 3)
When you solve the equation, you get that x = 0.5

The maximum volume is then:
(3 - 1)^2 * 0.5 = 2 units ^3

sahsjing:
My work was right.. just a little typo when copying it from paper to computer.. What I had originally written would not have given the correct answer.

Answer 2

You don't have enough information here. You don't even know if the cardboard is rectangular in shape, or some other shape., or even an irregular piece. Assuming that it's a square piece of cardboard, (the easiest case) here's how you'd do it ...
A 9 sq unit squrae piece of cardboard is 3 units long on each side. So if you cut away square sides of length s, each side would be of length and width (9 - 2s) and the volume would be (9-2s)^3. Differentiate, set to zero, and solve.
If the piece is not square, but rectangular, its dimensions are w by (9-w). Cutting away square corners of side s would give a box with dimensions (w-2s) by (9 - w - 2s) by s. Multiply these and get s(w-2s)(9-w-2s). Take the partial derivative with respect to s and solve. This would only give a solution in terms of w.
If it's an irregular shaped piece of cardboard the general solution of this problem would be a lot harder, it would involve idolving a differential equation to find the maximum dimensions of the cardboard then discarding parts of it to create a rectangle or square, which you would then solve as above.

Answer 3

If a and b are dimensions of cardboard, then its area s=ab=9; if we want to make a box with sides and bottom, then its volume v=x(a-2x)(b-2x), where x is height of the box or corner pieces to be scissored out; to find the max v we should find dv/dx first and then to demand dv/dx=0;
V=abx –2(a+b)x^2 +4x^3, v’=ab -4(a+b)x +12x^2=0, hence x= (2(a+b) - sqrt(4(a+b)^2 -12ab)) /12; if a=b=3, then x=(12 - sqrt(144-108)) /12 =0.5 units

Answer 4

Hmmn.If the 2 dimension area were to be folded into a cube then each side folded up would give an open cube. If u wanted a real cube u may have to take the cut pieces, join them after folding up the sides, and cover the sides to make that cube. All u may need is algebra.

Answer 5

remember that the formula for the fringe of a triangle is P = a + b + c yet, we've 30 inches of string, so 30 = a + b + c Heron's formula states that the part of any triangle is A = sqrt[S(S - a)(S - b)(S - c)], the place S = a million/2(a + b + c) In our case, considering a + b + c = 30, then S = (a million/2)(30) = 15, so A = sqrt[15(15 - a)(15 - b)(15 - c)] and that i admittedly have been given caught at this factor by using fact there is no way for me to specific A as one variable. i think of multivariable calculus is a technique or the different in touch, and that i'm no longer curious with reference to the respond myself.

Answer 6

Let x be the edge's length. Then we have
v = (x) (3-2x)^2

Solving dv/dx = 0 gives, x = 0.5, v(max) = 2 units^3

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Miss Linn,

Your work is not right even though you have the right answer.

From dV/dx = 0, you should get 9x-24x^2+12x^3 = 0, which can ba reduced into x(2x-1)(2x-3) = 0.