+279 Janine made a cylindrical vase in which the sum of the lateral area and area of one base was about 3000 square centimeters. The vase had a height of 50 centimeters. Find the radius of the vase. Explain the method you would use to find the radius
+128406 Actually...it's a pretty good question.....!!!
The surface area(s) are given by 2*pi*r*h (lateral SA) + pi*r^2 (SA of bottom) .... So we have.....
2*pi*r*h + pi*r^2 = 3000 And if h = 50, we have
2*pi*r*(50) + pi*r^2 = 3000
100*pi*r + pi*r^2 = 3000 Divide both sides by pi
100r + r^2 = 3000/pi Subtract 3000/pi from both sides
r^2 + 100r - 3000/pi = 0 Using the on-site calculator to solve, we have....
$${{\mathtt{r}}}^{{\mathtt{2}}}{\mathtt{\,\small\textbf+\,}}{\mathtt{100}}{\mathtt{\,\times\,}}{\mathtt{r}}{\mathtt{\,-\,}}{\frac{{\mathtt{3\,000}}}{{\mathtt{\pi}}}} = {\mathtt{0}} \Rightarrow \left\{ \begin{array}{l}{\mathtt{r}} = {\mathtt{\,-\,}}{\frac{\left({\mathtt{2}}{\mathtt{\,\times\,}}{{\mathtt{5}}}^{\left({\frac{{\mathtt{3}}}{{\mathtt{2}}}}\right)}{\mathtt{\,\times\,}}{\sqrt{{\mathtt{5}}{\mathtt{\,\times\,}}{{\mathtt{\pi}}}^{{\mathtt{2}}}{\mathtt{\,\small\textbf+\,}}{\mathtt{6}}{\mathtt{\,\times\,}}{\mathtt{\pi}}}}{\mathtt{\,\small\textbf+\,}}{\mathtt{50}}{\mathtt{\,\times\,}}{\mathtt{\pi}}\right)}{{\mathtt{\pi}}}}\\
{\mathtt{r}} = {\frac{\left({\mathtt{2}}{\mathtt{\,\times\,}}{{\mathtt{5}}}^{\left({\frac{{\mathtt{3}}}{{\mathtt{2}}}}\right)}{\mathtt{\,\times\,}}{\sqrt{{\mathtt{5}}{\mathtt{\,\times\,}}{{\mathtt{\pi}}}^{{\mathtt{2}}}{\mathtt{\,\small\textbf+\,}}{\mathtt{6}}{\mathtt{\,\times\,}}{\mathtt{\pi}}}}{\mathtt{\,-\,}}{\mathtt{50}}{\mathtt{\,\times\,}}{\mathtt{\pi}}\right)}{{\mathtt{\pi}}}}\\
\end{array} \right\} \Rightarrow \left\{ \begin{array}{l}{\mathtt{r}} = -{\mathtt{108.778\: \!649\: \!682\: \!953\: \!518\: \!7}}\\
{\mathtt{r}} = {\mathtt{8.778\: \!649\: \!682\: \!953\: \!518\: \!7}}\\
\end{array} \right\}$$
Reject the negative solution........so the radius is about 8.78 cm
+128406 Actually...it's a pretty good question.....!!!
The surface area(s) are given by 2*pi*r*h (lateral SA) + pi*r^2 (SA of bottom) .... So we have.....
2*pi*r*h + pi*r^2 = 3000 And if h = 50, we have
2*pi*r*(50) + pi*r^2 = 3000
100*pi*r + pi*r^2 = 3000 Divide both sides by pi
100r + r^2 = 3000/pi Subtract 3000/pi from both sides
r^2 + 100r - 3000/pi = 0 Using the on-site calculator to solve, we have....
$${{\mathtt{r}}}^{{\mathtt{2}}}{\mathtt{\,\small\textbf+\,}}{\mathtt{100}}{\mathtt{\,\times\,}}{\mathtt{r}}{\mathtt{\,-\,}}{\frac{{\mathtt{3\,000}}}{{\mathtt{\pi}}}} = {\mathtt{0}} \Rightarrow \left\{ \begin{array}{l}{\mathtt{r}} = {\mathtt{\,-\,}}{\frac{\left({\mathtt{2}}{\mathtt{\,\times\,}}{{\mathtt{5}}}^{\left({\frac{{\mathtt{3}}}{{\mathtt{2}}}}\right)}{\mathtt{\,\times\,}}{\sqrt{{\mathtt{5}}{\mathtt{\,\times\,}}{{\mathtt{\pi}}}^{{\mathtt{2}}}{\mathtt{\,\small\textbf+\,}}{\mathtt{6}}{\mathtt{\,\times\,}}{\mathtt{\pi}}}}{\mathtt{\,\small\textbf+\,}}{\mathtt{50}}{\mathtt{\,\times\,}}{\mathtt{\pi}}\right)}{{\mathtt{\pi}}}}\\
{\mathtt{r}} = {\frac{\left({\mathtt{2}}{\mathtt{\,\times\,}}{{\mathtt{5}}}^{\left({\frac{{\mathtt{3}}}{{\mathtt{2}}}}\right)}{\mathtt{\,\times\,}}{\sqrt{{\mathtt{5}}{\mathtt{\,\times\,}}{{\mathtt{\pi}}}^{{\mathtt{2}}}{\mathtt{\,\small\textbf+\,}}{\mathtt{6}}{\mathtt{\,\times\,}}{\mathtt{\pi}}}}{\mathtt{\,-\,}}{\mathtt{50}}{\mathtt{\,\times\,}}{\mathtt{\pi}}\right)}{{\mathtt{\pi}}}}\\
\end{array} \right\} \Rightarrow \left\{ \begin{array}{l}{\mathtt{r}} = -{\mathtt{108.778\: \!649\: \!682\: \!953\: \!518\: \!7}}\\
{\mathtt{r}} = {\mathtt{8.778\: \!649\: \!682\: \!953\: \!518\: \!7}}\\
\end{array} \right\}$$
Reject the negative solution........so the radius is about 8.78 cm
