Question

a. use your answers to parts c and f to express the new dimensions of the box.
b. the area of the bottom of the box is the product of the new length and width. write this area as a function in the form: a(x) = your answer
part iii: (8 points)
a. assume you want the bottom of your box to cover a total area of 16 in². set your question from part ii equal to 16 and use the quadratic formula to solve for the two possible answers.
b. one of these answers is not possible due to the original dimensions of the note card. clearly indicate which answer is not possible, and circle your final correct answer.

Explanation:

Step1: Since no information from parts c and f is given for part II - A, we'll focus on part III. First, assume the quadratic equation from part II is of the form $ax^{2}+bx + c=0$.

Let the quadratic equation be $ax^{2}+bx + c = 0$. The quadratic formula is $x=\frac{-b\pm\sqrt{b^{2}-4ac}}{2a}$.

Step2: For part III - A, set the area function $A(x)$ (from part II - B) equal to 16. So we have $ax^{2}+bx + c=16$, which can be rewritten as $ax^{2}+bx+(c - 16)=0$. Then substitute $a$, $b$, and $c$ into the quadratic formula $x=\frac{-b\pm\sqrt{b^{2}-4a(c - 16)}}{2a}$ to find the two possible values of $x$.

Step3: For part III - B, consider the physical constraints based on the original dimensions of the note - card. If one of the solutions for $x$ makes the length or width of the box negative (or violates some other non - negative or physical constraints), then that solution is not possible.

Answer:

A. The two possible values of $x$ are found using $x=\frac{-b\pm\sqrt{b^{2}-4a(c - 16)}}{2a}$ (after finding the quadratic equation from setting $A(x)=16$).
B. The non - possible solution is the one that violates the physical constraints of the original note - card dimensions. The correct solution is the other value of $x$.