Answer:
Hence, to maximize the volume of the open box, the side length of the squares cut out from the corners of the cardboard should be 9 units.
x=9
Step-by-step explanation:
To find the value of x that maximizes the volume of the open box made from cutting squares out of a square cardboard, we can use optimization techniques.
Let's assume that x is the side length of the squares cut out from the corners of the cardboard.
The dimensions of the resulting open box will be (12-2x) x (12-2x) x x, where 12-2x is the length and width of the base, and x is the height.
To find the volume of the box, we multiply these dimensions:
V = (12-2x) * (12-2x) * x
To maximize the volume, we can take the derivative of V with respect to x, set it equal to zero, and solve for x.
dV/dx = 0
Differentiating V with respect to x, we get:
dV/dx = (12-2x) * (12-2x) + 2 * (12-2x) * (-2) * x + (12-2x) * x
Simplifying the expression, we have:
dV/dx = 4x^2 - 48x + 144 + 48x - 8x^2 + 12x
Combining like terms, we get:
dV/dx = -4x^2 + 60x + 144
Setting dV/dx equal to zero and solving for x:
-4x^2 + 60x + 144 = 0
Simplifying the equation, we have:
x^2 - 15x - 36 = 0
Factoring the equation, we get:
(x - 9)(x + 4) = 0
Solving for x, we have two possible solutions: x = 9 and x = -4.
Since x represents the side length of a square, it cannot be negative. Therefore, x = 9.