question about find max value...?

Question

1. An open box is formed by a rectangular sheet of cardboard by cutting equal squares from each corner and folding up the edges. If the dimensions of the carboard are 15 inches by 24 inches, what size squres should be cut out to obtain a box of maximum volume?

2. Suppose a rectangle has it lovwer base on the x-axis and upper vertices on the graph of the function y=8-x^2. Find the area of the largest such rectangle.

Answer 1

1. The initial dimensions of the cardboard before cutting are 15 x 24. We cut out equal squares of side x, leaving

l = 24 - 2x
w = 15 - 2x
h = x

the volume is

V = (24 - 2x)(15 - 2x)x
= 4x^3 - 78x^2 + 360x

Leave us take the first derivative of this, to get the maximum:

0 = 12x^2 - 156x + 360
= x^2 - 13x + 30
= (x - 10)(x - 3)

The solution x = 10 is not reasonable, since it will leave us with negative width. Using x = 3, we get

l = 18
w = 9
h = 3
V = 486

2. This is a downward-opening parabola that has the y-axis as its axis of symmetry and a maximum at (0, 8). HayHrbr's analysis is correct, and the first derivative of the area gives us

16 = 12x^2
x^2 = 4/3
x = 2 / sqrt(3)
= 2sqrt(3) / 3

Answer 2

Is this calculus where you use first derivative to get maximum? If so draw the figure to see that the volume would be x(15-2x)(24-2x)
and find the derivative, solve it for 0.

Draw the rectangle. Label the bottom 2 vertices (x,0) and (-x,0). Then the top two vertices would be (x, 8-x^2) and (-x, 8-x^2). So it has a width of 2x and a height of (8-x^2) so its area is
16x - 4x^3. Again, find dA/Dx, set it equal to 0, solve, etc

Answer 3

Do you have a TI 83 (or 84)calculator?
I don't think its possible without one (unless you want to physically figure it out)
If you do
type the equation in Y=
then graph
the click second then calc
go down to maximum
psoition the cursor to the left of the max point and press enter
then to the right and enter
click enter again
it should show it