+3693 A box is to be formed by cutting square pieces out of the corner of a rectangular piece of a 3" x 5" notecard as shown below. The sides are then folded up to form a box.
a)Write the function that expresses the area of the bottom of the box as a function of the length of the side of one of the square pieces. (6 points)
b)How large should x be in order for the area of the bottom of the box to equal 10 in2? Round your answer to the nearest hundredth.
Thank you!!!
+128475
a)......Since we're cutting two "x"s" off each side, the area of the bottom of the box is just W * L = (5 - 2x) * (3-2x)
b) So we have
(5 - 2x)(3 -2x) = 10 simplify
15 - 16x + 4x^2 = 10 rearrange
4x^2 - 16x + 5 = 0
Using the onsite solver (since this doesn't factor), we have
$${\mathtt{4}}{\mathtt{\,\times\,}}{{\mathtt{x}}}^{{\mathtt{2}}}{\mathtt{\,-\,}}{\mathtt{16}}{\mathtt{\,\times\,}}{\mathtt{x}}{\mathtt{\,\small\textbf+\,}}{\mathtt{5}} = {\mathtt{0}} \Rightarrow \left\{ \begin{array}{l}{\mathtt{x}} = {\mathtt{\,-\,}}{\frac{\left({\sqrt{{\mathtt{11}}}}{\mathtt{\,-\,}}{\mathtt{4}}\right)}{{\mathtt{2}}}}\\
{\mathtt{x}} = {\frac{\left({\sqrt{{\mathtt{11}}}}{\mathtt{\,\small\textbf+\,}}{\mathtt{4}}\right)}{{\mathtt{2}}}}\\
\end{array} \right\} \Rightarrow \left\{ \begin{array}{l}{\mathtt{x}} = {\mathtt{0.341\: \!687\: \!604\: \!822\: \!300\: \!1}}\\
{\mathtt{x}} = {\mathtt{3.658\: \!312\: \!395\: \!177\: \!699\: \!9}}\\
\end{array} \right\}$$
Reject the larger answer...(it would make the length of both sides negative).....
.34 (rounded) "works"
+128475
a)......Since we're cutting two "x"s" off each side, the area of the bottom of the box is just W * L = (5 - 2x) * (3-2x)
b) So we have
(5 - 2x)(3 -2x) = 10 simplify
15 - 16x + 4x^2 = 10 rearrange
4x^2 - 16x + 5 = 0
Using the onsite solver (since this doesn't factor), we have
$${\mathtt{4}}{\mathtt{\,\times\,}}{{\mathtt{x}}}^{{\mathtt{2}}}{\mathtt{\,-\,}}{\mathtt{16}}{\mathtt{\,\times\,}}{\mathtt{x}}{\mathtt{\,\small\textbf+\,}}{\mathtt{5}} = {\mathtt{0}} \Rightarrow \left\{ \begin{array}{l}{\mathtt{x}} = {\mathtt{\,-\,}}{\frac{\left({\sqrt{{\mathtt{11}}}}{\mathtt{\,-\,}}{\mathtt{4}}\right)}{{\mathtt{2}}}}\\
{\mathtt{x}} = {\frac{\left({\sqrt{{\mathtt{11}}}}{\mathtt{\,\small\textbf+\,}}{\mathtt{4}}\right)}{{\mathtt{2}}}}\\
\end{array} \right\} \Rightarrow \left\{ \begin{array}{l}{\mathtt{x}} = {\mathtt{0.341\: \!687\: \!604\: \!822\: \!300\: \!1}}\\
{\mathtt{x}} = {\mathtt{3.658\: \!312\: \!395\: \!177\: \!699\: \!9}}\\
\end{array} \right\}$$
Reject the larger answer...(it would make the length of both sides negative).....
.34 (rounded) "works"
