an open box is to be made from a square piece of cardboard whose sides are 19 inches long,by cutting squares of equal size from the corners and bending up the sides. Determine the size of the aquare that is to be cut out so that the volume may be a maximum.
Let the size of the cut-out square be x inches
so the base of the box is (19-2x) by (19-2x) and the height is x
V = x(19-2x)^2
= x(361 - 76x + 4x^2)
= 4x^3 - 76x^2 + 361x
dV/dx = 12x^2 - 152x + 361
= 0 for a max of V
x = (152 ± √5776)/24
= (152 ± 76)/24 = 19/2 or 19/6
but 0 < x < 19/2
so x = 19/6
the square to be cut out should be appr 3.17 inches by 3.17 inches.
(notice our quadratic 12x^2 - 152x + 361 = 0 factors to
(2x - 19)(6x - 19) = 0
x = 19/2 or x = 19/6
To find the size of the square that needs to be cut out to maximize the volume of the open box, we need to consider the dimensions of the box and its volume.
Let's call the side length of the square to be cut out as x inches. By cutting a square of size x from each corner of the original square cardboard, the side length of the resulting box will be (19 - 2x) inches.
The height of the box will be x inches, as we are bending up the sides of the cardboard.
Now, the volume of the box can be calculated by multiplying its width, length, and height. In this case, the width and length will be the same, given by (19 - 2x).
So, the volume V of the box can be expressed as:
V = (19 - 2x) * (19 - 2x) * x
To maximize the volume, we need to find the value of x that yields the maximum V.
To do this, we can use calculus. We differentiate V with respect to x, set it equal to zero, and solve for x.
dV/dx = 4x^2 - 76x + 361 = 0
Now, you can solve the quadratic equation to find the value(s) of x that satisfy this equation. The value of x that is positive and less than (19/2) will be the one that maximizes the volume.
Once you find this value of x, you can substitute it back into the equation (19 - 2x) to find the side length of the square that needs to be cut out to maximize the volume of the box.