the figure shows a square sheet of cardboard on the left and an open - box on the right. the square sheet…

the 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. 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). a 2 b 3 c 4 d 6

the 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. 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). a 2 b 3 c 4 d 6

Answer

Explanation:

Step1: Find volume formula

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

Step2: Differentiate the volume function

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

Step3: Set the derivative equal to zero

$12x^{2}-96x + 144 = 0$. Divide through by 12: $x^{2}-8x + 12=0$.

Step4: Solve the quadratic equation

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 possible for the dimensions of a box. So we discard $x = 6$.

Answer:

A. 2