a box with an open top is to be made from a rectangular piece of tin by cutting equal squares from the corners and turning up the sides. The piece of tin measures 1mx2m. Find the size of the squares that yields a maximum capacity for the box.
So far i have
V=(1-2x)(2-2x)x
so, figure dV/dx
dV/dx = 2(6x^2-6x+1)
dV/dx = 0 when x = 1/6 (3±√3) = .211 or .789
Now .789 is impossible, since the width is only 1.
so, the cuts are .211m
how did you go from 0= 1/6 (3+-�ã3)
dV/dx = 2(6x^2-6x+1
so, dV/dx = 0 when 6x^2-6x+1
solve the quadratic to get X = 1/6 (3±√3)
this is calculus; algebra I should be no problem...
To find the size of the squares that yields a maximum capacity for the box, you need to find the value of x that maximizes the volume equation V = (1 - 2x)(2 - 2x)x.
Let's go through the steps to solve this problem:
1. Start with the equation for the volume of the box: V = (1 - 2x)(2 - 2x)x.
2. Expand the equation: V = (2 - 4x - 2x + 4x^2)x.
3. Simplify: V = 2x - 8x^2 + 4x^3.
4. Rearrange the equation: V = 4x^3 - 8x^2 + 2x.
5. To find the maximum volume, take the derivative of V with respect to x and set it equal to zero: dV/dx = 12x^2 - 16x + 2 = 0.
6. Solve the quadratic equation: 12x^2 - 16x + 2 = 0. You can use the quadratic formula or factoring to solve for x. In this case, factoring is easier: (2x - 1)(6x - 2) = 0. So x = 1/2 or x = 1/6.
7. Since we're dealing with dimensions of squares, we discard x = 1/6 since it is too small to form a box.
8. Finally, substitute x = 1/2 back into the volume equation to find the maximum volume: V = 4 * (1/2)^3 - 8 * (1/2)^2 + 2 * (1/2) = 1 cubic meter.
Therefore, the size of the squares that yields a maximum capacity for the box is 1/2 meters.