Prove the limit statement
-You are asked to make some circular cylinders, each with a cross-sectional area of $9 \mathrm {~cm} ^ { 2 }$. To do this, you need to know how much deviation from the ideal cylinder diameter of $x _ { 0 } = 1.93 \mathrm {~cm}$ you can allow and still have the area come within $0.1 \mathrm {~cm} ^ { 2 }$ of the required $9 \mathrm {~cm} ^ { 2 }$. To find out, let $A = \pi \left( \frac { x } { 2 } \right) ^ { 2 }$ and look for the interval in which you must hold $x$ to make $| \mathrm { A } - 9 |

Question

Prove the limit statement
-You are asked to make some circular cylinders, each with a cross-sectional area of $9 \mathrm {~cm} ^ { 2 }$. To do this, you need to know how much deviation from the ideal cylinder diameter of $x _ { 0 } = 1.93 \mathrm {~cm}$ you can allow and still have the area come within $0.1 \mathrm {~cm} ^ { 2 }$ of the required $9 \mathrm {~cm} ^ { 2 }$. To find out, let $A = \pi \left( \frac { x } { 2 } \right) ^ { 2 }$ and look for the interval in which you must hold $x$ to make $| \mathrm { A } - 9 | < 0.1$. What interval do you find?
A) $( 3.3663,3.4039 )$
B) $( 0.5642,0.5642 )$
C) $( 2.3803,2.4069 )$
D) $( 5.9666,6.0332 )$

Quizplus · Accepted Answer

Given $A = \pi \left( \frac{x}{2} \right)^2$ and $|A - 9| < 0.1$, we aim to find the interval for $x$ that satisfies this condition. The area of a circle is given by $A = \pi r^2$, where $r$ is the radius. In this case, $r = \frac{x}{2}$, so $A = \pi \left( \frac{x}{2} \right)^2$. The target area is $9 \mathrm{cm}^2$, so we solve for $x$ when $A$ is slightly less than $9.1 \mathrm{cm}^2$ and slightly more than $8.9 \mathrm{cm}^2$. Solving these equations gives us the interval for $x$ that keeps the area within the desired range, leading to the correct interval for $x$.

Prove the Limit Statement -You Are Asked to Make Some 9 cm29 \mathrm {~cm} ^ { 2 }9 cm2

Prove the Limit Statement
-You Are Asked to Make Some $9 \mathrm {~cm} ^ { 2 }$