practice exam 3 mcq, part b - calculator required\nthe figure shows a square sheet of cardboard on the left…

practice exam 3 mcq, part b - calculator required\nthe figure shows a square sheet of cardboard on the left and an open - box on the right. the square sheet has side length 12 inches (in) and is used to make the open box by removing from each of the four corners a square of side length (x) inches, and folding up the sides.\n9. for what value of (x) does the box have the maximum possible volume? note: the volume of a rectangular solid with length (l), width (w), and height (h) is (v = lwh).\n(a) 2\n(b) 3\n(c) 4\n(d) 6\nin the polar coordinate system, the graph of a polar function (r = f(\theta)) is shown with a domain of all real values of (\theta) for (0leq\theta<2pi). on this interval of (\theta), the graph has no holes, passes through each point exactly one time, and as (\theta) increases, the graph passes through the labeled points (a), (b), (c), and (d), in that order.\n10. on which of the following intervals is the average rate of change of (r) with respect to (\theta) greatest?\n(a) from (a) to (c)\n(b) from (b) to (c)\n(c) from (b) to (d)\n(d) from (c) to (d)

practice exam 3 mcq, part b - calculator required\nthe figure shows a square sheet of cardboard on the left and an open - box on the right. the square sheet has side length 12 inches (in) and is used to make the open box by removing from each of the four corners a square of side length (x) inches, and folding up the sides.\n9. for what value of (x) does the box have the maximum possible volume? note: the volume of a rectangular solid with length (l), width (w), and height (h) is (v = lwh).\n(a) 2\n(b) 3\n(c) 4\n(d) 6\nin the polar coordinate system, the graph of a polar function (r = f(\theta)) is shown with a domain of all real values of (\theta) for (0leq\theta<2pi). on this interval of (\theta), the graph has no holes, passes through each point exactly one time, and as (\theta) increases, the graph passes through the labeled points (a), (b), (c), and (d), in that order.\n10. on which of the following intervals is the average rate of change of (r) with respect to (\theta) greatest?\n(a) from (a) to (c)\n(b) from (b) to (c)\n(c) from (b) to (d)\n(d) from (c) to (d)

Answer

Explanation:

Step1: Find the volume function

The length and width of the box after cutting and folding are $l = w=12 - 2x$, and the height is $h=x$. So the volume function $V(x)=(12 - 2x)(12 - 2x)x=(12 - 2x)^2x=x(144-48x + 4x^{2})=4x^{3}-48x^{2}+144x$.

Step2: Find the derivative of the volume function

Using the power - rule $(x^n)'=nx^{n - 1}$, we have $V'(x)=12x^{2}-96x + 144$.

Step3: Set the derivative equal to zero and solve for $x$

Set $V'(x)=0$, so $12x^{2}-96x + 144 = 0$. Divide through by 12: $x^{2}-8x + 12=0$. Factor the quadratic equation: $(x - 2)(x - 6)=0$. So $x = 2$ or $x = 6$. But if $x = 6$, then $12-2x=0$, which is not valid for the dimensions of a box. So we consider $x = 2$.

Step4: Check the second - derivative to confirm it's a maximum

Find the second - derivative $V''(x)=24x-96$. When $x = 2$, $V''(2)=24\times2-96=48 - 96=-48<0$. So the volume is maximized at $x = 2$.

Answer:

A. 2