an open box is to be made by cutting small congruent squares from corners of a 12cm by 12cm . sheet of tin and bending up the sides . how large should the squares cut from the corners to be make the box hold as much as possible ?
no
Let the Volume be V, and side of the square to be cut be x
V=x(12-x)^2
V=144x-48x^2+4x^3
dv/dx=144-96x+12x^2=0
12x^2-96x+144=0
x^2-8x+12=0
Solve for x
(x-2)(x-6)=0
x=2 0r x=6
To solve this problem, we need to determine the dimensions of the cut that will maximize the volume of the resulting box.
Let's consider the dimensions of the box after the corners are cut and the sides are folded up. Since we are cutting congruent squares from the corners, each side of the resulting box will have a length of (12cm - 2x), where "x" represents the size of a square cut from the corners.
To find the volume of the box, we multiply the length, width, and height. In this case, the length and width are equal to (12cm - 2x), and the height is equal to "x".
Therefore, the volume of the box is V = (12cm - 2x) * (12cm - 2x) * x.
To find the value of "x" that maximizes V, we can use calculus by taking the derivative of V with respect to "x" and setting it to zero, then solving for "x".
Taking the derivative of V, we get:
dV/dx = 4(12cm - 2x)(x) - (12cm - 2x)(12cm - 2x)
= 48x - 16x^2 - 144cm + 24x^2 + 48x
= 48x - 16x^2 + 24x^2 + 48x - 144cm
= 8x^2 + 96x - 144cm
Now, setting dV/dx to zero and solving for "x":
8x^2 + 96x - 144cm = 0
We can simplify the equation by dividing through by 8:
x^2 + 12x - 18cm = 0
This is a quadratic equation that we can solve using the quadratic formula:
x = (-b ± √(b^2 - 4ac)) / (2a)
For our equation, a = 1, b = 12, and c = -18cm. Plugging in these values, we get:
x = (-12 ± √(12^2 - 4(1)(-18cm))) / (2(1))
Simplifying further:
x = (-12 ± √(144 + 72cm)) / 2
x = (-12 ± √(144(1 + 0.5cm))) / 2
x = (-12 ± √(144 + 72cm)) / 2
x = (-12 ± √(144(1 + 0.5cm))) / 2
x = (-12 ± 12√(1 + 0.5cm)) / 2
x = -6 ± 6√(1 + 0.5cm)
The value of "x" that maximizes the volume of the box should be the positive root, so:
x = -6 + 6√(1 + 0.5cm)
This means that the size of each square cut from the corners should be approximately -6 + 6√(1 + 0.5cm) cm to maximize the volume of the box. Since a negative length doesn't make sense in this context, we can ignore the negative root and only consider the positive root.
Please note that the above answer assumes that the sides of the box will be square when folded up. If you want different dimensions or a rectangular box, the approach and calculations will differ.