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 we don't have answers from parts c and f for part II - A, we'll focus on part III. Let the area function from part II be a quadratic function of the form $A(x)=ax^{2}+bx + c$. Set $A(x) = 16$, so we have the quadratic equation $ax^{2}+bx + c-16=0$.

The quadratic formula for a quadratic equation $Ax^{2}+Bx + C = 0$ is $x=\frac{-B\pm\sqrt{B^{2}-4AC}}{2A}$.

Step2: Identify $A = a$, $B = b$, $C=c - 16$ in our equation $ax^{2}+bx+(c - 16)=0$ and substitute into the quadratic formula to get $x=\frac{-b\pm\sqrt{b^{2}-4a(c - 16)}}{2a}$. Calculate the two values of $x$.

Step3: For part III - B, check the two values of $x$ obtained in Step2 against the original dimensions of the note - card. If a value of $x$ makes the new length or width of the box negative (or violates some other physical constraint based on the original note - card dimensions), then that value is not possible. Circle the valid value.

Answer:

The two possible values of $x$ from part III - A are $x_1=\frac{-b+\sqrt{b^{2}-4a(c - 16)}}{2a}$ and $x_2=\frac{-b-\sqrt{b^{2}-4a(c - 16)}}{2a}$. After checking against the original note - card dimensions in part III - B, circle the valid value. (Note: Without the actual function $A(x)$ from part II, we can't give numerical answers).