Question To meet the required cylinder area of $11 \mathrm{in}^{2}$ within $0.01 \mathrm{in}^{2}$ , the cylinder diameter $x$ must be held in the interval $[3.730, 3.754]$ inches.

Studdy Solution

STEP 1

Assumptions
1. The ideal cylinder diameter is $c = 3.742$ inches.
2. The cross-sectional area $A$ of the cylinder is given by the formula $A = \pi \left(\frac{x}{2}\right)^2$ , where $x$ is the diameter of the cylinder.
3. The required cross-sectional area is $11 \mathrm{in}^2$ .
4. The allowable deviation in area is $0.01 \mathrm{in}^2$ .
5. We need to find the interval for $x$ such that the area $A$ is within $0.01 \mathrm{in}^2$ of the required area.

STEP 2

First, we need to establish the range for the area $A$ that is acceptable. This range is from $11 - 0.01$ to $11 + 0.01$ .
$11 - 0.01 \leq A \leq 11 + 0.01$

STEP 3

Now, we substitute the formula for $A$ into the inequality.
$11 - 0.01 \leq \pi \left(\frac{x}{2}\right)^2 \leq 11 + 0.01$

STEP 4

Simplify the inequality by subtracting $11$ from all parts of the inequality.
$-0.01 \leq \pi \left(\frac{x}{2}\right)^2 - 11 \leq 0.01$

STEP 5

Divide all parts of the inequality by $\pi$ to isolate the term with $x$ .
$\frac{-0.01}{\pi} \leq \left(\frac{x}{2}\right)^2 - \frac{11}{\pi} \leq \frac{0.01}{\pi}$

STEP 6

Add $\frac{11}{\pi}$ to all parts of the inequality to isolate the squared term.
$\frac{-0.01}{\pi} + \frac{11}{\pi} \leq \left(\frac{x}{2}\right)^2 \leq \frac{0.01}{\pi} + \frac{11}{\pi}$

STEP 7

Calculate the numerical values for the left and right endpoints of the inequality.
$\frac{-0.01}{\pi} + \frac{11}{\pi} \approx 3.4784$ $\frac{0.01}{\pi} + \frac{11}{\pi} \approx 3.5016$

STEP 8

Now, we take the square root of all parts of the inequality to solve for $x$ .
$\sqrt{3.4784} \leq \frac{x}{2} \leq \sqrt{3.5016}$

STEP 9

Multiply all parts of the inequality by $2$ to solve for $x$ .
$2\sqrt{3.4784} \leq x \leq 2\sqrt{3.5016}$

STEP 10

Calculate the numerical values for $x$ .
$2\sqrt{3.4784} \approx 3.733$ $2\sqrt{3.5016} \approx 3.751$

STEP 11

Round the left endpoint up and the right endpoint down to the nearest thousandth as needed.
$x \approx [3.733, 3.751]$
Therefore, the diameter $x$ must be held within the interval $[3.733, 3.751]$ inches to meet the requirements for the area.

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Question To meet the required cylinder area of $11 \mathrm{in}^{2}$ within $0.01 \mathrm{in}^{2}$ , the cylinder diameter $x$ must be held in the interval $[3.730, 3.754]$ inches.

Studdy Solution

STEP 1

STEP 2

First, we need to establish the range for the area $A$ that is acceptable. This range is from $11 - 0.01$ to $11 + 0.01$ .
$11 - 0.01 \leq A \leq 11 + 0.01$

STEP 3

Now, we substitute the formula for $A$ into the inequality.
$11 - 0.01 \leq \pi \left(\frac{x}{2}\right)^2 \leq 11 + 0.01$

STEP 4

Simplify the inequality by subtracting $11$ from all parts of the inequality.
$-0.01 \leq \pi \left(\frac{x}{2}\right)^2 - 11 \leq 0.01$

STEP 5

Divide all parts of the inequality by $\pi$ to isolate the term with $x$ .
$\frac{-0.01}{\pi} \leq \left(\frac{x}{2}\right)^2 - \frac{11}{\pi} \leq \frac{0.01}{\pi}$

STEP 6

Add $\frac{11}{\pi}$ to all parts of the inequality to isolate the squared term.
$\frac{-0.01}{\pi} + \frac{11}{\pi} \leq \left(\frac{x}{2}\right)^2 \leq \frac{0.01}{\pi} + \frac{11}{\pi}$

STEP 7

Calculate the numerical values for the left and right endpoints of the inequality.
$\frac{-0.01}{\pi} + \frac{11}{\pi} \approx 3.4784$ $\frac{0.01}{\pi} + \frac{11}{\pi} \approx 3.5016$

STEP 8

Now, we take the square root of all parts of the inequality to solve for $x$ .
$\sqrt{3.4784} \leq \frac{x}{2} \leq \sqrt{3.5016}$

STEP 9

Multiply all parts of the inequality by $2$ to solve for $x$ .
$2\sqrt{3.4784} \leq x \leq 2\sqrt{3.5016}$

STEP 10

Calculate the numerical values for $x$ .
$2\sqrt{3.4784} \approx 3.733$ $2\sqrt{3.5016} \approx 3.751$

STEP 11

Round the left endpoint up and the right endpoint down to the nearest thousandth as needed.
$x \approx [3.733, 3.751]$
Therefore, the diameter $x$ must be held within the interval $[3.733, 3.751]$ inches to meet the requirements for the area.

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Question To meet the required cylinder area of 11in211 \mathrm{in}^{2}11in2 within 0.01in20.01 \mathrm{in}^{2}0.01in2, the cylinder diameter xxx must be held in the interval [3.730,3.754][3.730, 3.754][3.730,3.754] inches.

Studdy Solution

STEP 1

STEP 2

First, we need to establish the range for the area A A A that is acceptable. This range is from 11−0.01 11 - 0.01 11−0.01 to 11+0.01 11 + 0.01 11+0.01.11−0.01≤A≤11+0.0111 - 0.01 \leq A \leq 11 + 0.0111−0.01≤A≤11+0.01

STEP 3

Now, we substitute the formula for A A A into the inequality.11−0.01≤π(x2)2≤11+0.0111 - 0.01 \leq \pi \left(\frac{x}{2}\right)^2 \leq 11 + 0.0111−0.01≤π(2x​)2≤11+0.01

STEP 4

Simplify the inequality by subtracting 11 11 11 from all parts of the inequality.−0.01≤π(x2)2−11≤0.01-0.01 \leq \pi \left(\frac{x}{2}\right)^2 - 11 \leq 0.01−0.01≤π(2x​)2−11≤0.01

STEP 5

Divide all parts of the inequality by π \pi π to isolate the term with x x x.−0.01π≤(x2)2−11π≤0.01π\frac{-0.01}{\pi} \leq \left(\frac{x}{2}\right)^2 - \frac{11}{\pi} \leq \frac{0.01}{\pi}π−0.01​≤(2x​)2−π11​≤π0.01​

STEP 6

Add 11π \frac{11}{\pi} π11​ to all parts of the inequality to isolate the squared term.−0.01π+11π≤(x2)2≤0.01π+11π\frac{-0.01}{\pi} + \frac{11}{\pi} \leq \left(\frac{x}{2}\right)^2 \leq \frac{0.01}{\pi} + \frac{11}{\pi}π−0.01​+π11​≤(2x​)2≤π0.01​+π11​

STEP 7

Calculate the numerical values for the left and right endpoints of the inequality.−0.01π+11π≈3.4784\frac{-0.01}{\pi} + \frac{11}{\pi} \approx 3.4784π−0.01​+π11​≈3.4784 0.01π+11π≈3.5016\frac{0.01}{\pi} + \frac{11}{\pi} \approx 3.5016π0.01​+π11​≈3.5016

STEP 8

Now, we take the square root of all parts of the inequality to solve for x x x.3.4784≤x2≤3.5016\sqrt{3.4784} \leq \frac{x}{2} \leq \sqrt{3.5016}3.4784​≤2x​≤3.5016​

STEP 9

Multiply all parts of the inequality by 2 2 2 to solve for x x x.23.4784≤x≤23.50162\sqrt{3.4784} \leq x \leq 2\sqrt{3.5016}23.4784​≤x≤23.5016​

STEP 10

Calculate the numerical values for x x x.23.4784≈3.7332\sqrt{3.4784} \approx 3.73323.4784​≈3.733 23.5016≈3.7512\sqrt{3.5016} \approx 3.75123.5016​≈3.751

STEP 11

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Question To meet the required cylinder area of $11 \mathrm{in}^{2}$ within $0.01 \mathrm{in}^{2}$ , the cylinder diameter $x$ must be held in the interval $[3.730, 3.754]$ inches.

First, we need to establish the range for the area $A$ that is acceptable. This range is from $11 - 0.01$ to $11 + 0.01$ .
$11 - 0.01 \leq A \leq 11 + 0.01$

Now, we substitute the formula for $A$ into the inequality.
$11 - 0.01 \leq \pi \left(\frac{x}{2}\right)^2 \leq 11 + 0.01$

Simplify the inequality by subtracting $11$ from all parts of the inequality.
$-0.01 \leq \pi \left(\frac{x}{2}\right)^2 - 11 \leq 0.01$

Divide all parts of the inequality by $\pi$ to isolate the term with $x$ .
$\frac{-0.01}{\pi} \leq \left(\frac{x}{2}\right)^2 - \frac{11}{\pi} \leq \frac{0.01}{\pi}$

Add $\frac{11}{\pi}$ to all parts of the inequality to isolate the squared term.
$\frac{-0.01}{\pi} + \frac{11}{\pi} \leq \left(\frac{x}{2}\right)^2 \leq \frac{0.01}{\pi} + \frac{11}{\pi}$

Calculate the numerical values for the left and right endpoints of the inequality.
$\frac{-0.01}{\pi} + \frac{11}{\pi} \approx 3.4784$ $\frac{0.01}{\pi} + \frac{11}{\pi} \approx 3.5016$

Now, we take the square root of all parts of the inequality to solve for $x$ .
$\sqrt{3.4784} \leq \frac{x}{2} \leq \sqrt{3.5016}$

Multiply all parts of the inequality by $2$ to solve for $x$ .
$2\sqrt{3.4784} \leq x \leq 2\sqrt{3.5016}$

Calculate the numerical values for $x$ .
$2\sqrt{3.4784} \approx 3.733$ $2\sqrt{3.5016} \approx 3.751$