Pre Calculus Problem. Bet you can't do it!?

Question

It's one of the hardest problems in Pre-cal history !

A wire 6 meters long is cut into 12 pieces: eight of one length and four of another. These pieces are welded together at right angles to form the frame of a box with a square base.How long should each wire piece be to maximize the volume of the box? How long should each wire be to maximize the total surface area of the box?

I attempted this problem like 10 times already!

Answer 1

8 pieces each x m (square base and height)
4 pieces each y m (vertical edges)
8x + 4y = 6
4x + 2y = 3
y = 3/2 - 2x
V(x) = x² . (3/2 - 2x)
V(x) = (3/2).x² - 2x³
V `(x) = 3x - 6x² = 3x.(1 - 2x) = 0 for MAX.
x = 1 / 2 for max. volume.
y = 3/2 - 1 = 1/2
8 pieces at 0.5 m
4 pieces at 0.5 m
ie 12 pieces at 0.5 m (i.e a cube)

Answer 2

First of all, we have:
8x+4y = 6, where x and y represent the side lengths we seek. The box has a square frame, and visualizing this 3d object as a rectangular box, we can see that the volume is given by: V=yx^2.
It must be this; we have eight sticks of length x, and 4 of length y. Draw a picture and you will see.
But from above, we have a relationship between x and y forced from the length of the wire. Thus you can sub the first equation into the second. First, solve for y in the first equation:
y = 1.5 - 2x
Now sub that into the second equation to get:
V=(1.5-2x)x^2
Here's where you need a little calculus. To find the extremums of any function, take the derivative and set it equal to zero. In this case we need (dV/dx) = 0 :
(dV/dx) = (-2)x^2 + (1.5-2x)2x
= -6x^2 + 3x
Now set this equal to zero to find x=0 and 0.5. Clearly x=0 is an extremum we're not interested in, so x=0.5 is the answer. Looking back at the first equation, you'll see that y=0.5 as well and we have a cube.

The area will be given by:
A = 2x^2 + 4xy
Now sub in the first equation to get:
A = -6x^2 +6x
Now we need (dA/dx) = 0 :
-12x + 6 = 0
which yields x=0.5, and you get a cube again!

Note that you must draw a picture with the given constraints to understand this problem.

Answer 3

So, you would have 8 pieces of length x, and 4 pieces of length ((6-8x)/4). The area of each side is (x)(x), (x)(6-8x)/4, and (x)(6-8x)/4. The total surface area is then S(x)= 2(x)(x) + 4(x)(6-8x)/4 = 2x^2 + 6x - 8x^2 = -6x^2 +6x. The use a calculator, calculus, or the formula for the vertex of a parabola to find that the maximum occurs at x= .5. So, the maximum surface area is when the dimensions are .5m by .5m by .5m, which gives a surface area of 1.5 square meters.

Answer 4

im thinking that 6 meters = 18 ft
so the four pieces would be 2 1/2 ft long
and the eight pieces would be 1 ft long

Answer 5

f(x) = x(x-4)(x-2)(x+2(x+4) So if ok = 0, f(x) has 5 roots: -4,-2.0,2,4 ok = 30 ends up in basically a million fee ok = 20 ends up in 3 roots there is not any fee of ok that ends up in no zeroes this could be a 5th order equation and has 5 roots. the main imaginarr roots it could have is 4 with the aid of fact they continuously are available in conjugate pairs. subsequently there could desire to be one real root.