Westonci.ca connects you with experts who provide insightful answers to your questions. Join us today and start learning! Join our Q&A platform and get accurate answers to all your questions from professionals across multiple disciplines. Experience the convenience of finding accurate answers to your questions from knowledgeable experts on our platform.
Sagot :
To solve the problem of finding the radius [tex]\( r \)[/tex] that minimizes the total surface area [tex]\( S \)[/tex] of the cylindrical container, we'll follow a systematic approach. Given the total surface area function:
[tex]\[ S = 2 \pi r^2 + \frac{100}{r} \][/tex]
we need to find the radius [tex]\( r \)[/tex] that minimizes this function. Let's break this down step-by-step:
### Step 1: Define the Surface Area Function
The surface area [tex]\( S \)[/tex] is a function of the radius [tex]\( r \)[/tex]:
[tex]\[ S(r) = 2 \pi r^2 + \frac{100}{r} \][/tex]
### Step 2: Find the First Derivative
To find the critical points where [tex]\( S \)[/tex] might be minimized, we need to find the first derivative of [tex]\( S \)[/tex] with respect to [tex]\( r \)[/tex]:
[tex]\[ S'(r) = \frac{d}{dr} \left( 2 \pi r^2 + \frac{100}{r} \right) \][/tex]
Using the power rule and the quotient rule, we get:
[tex]\[ S'(r) = 4 \pi r - \frac{100}{r^2} \][/tex]
### Step 3: Solve for Critical Points
Set the first derivative equal to zero and solve for [tex]\( r \)[/tex]:
[tex]\[ 4 \pi r - \frac{100}{r^2} = 0 \][/tex]
This can be rearranged to:
[tex]\[ 4 \pi r = \frac{100}{r^2} \][/tex]
Multiply both sides by [tex]\( r^2 \)[/tex] to get rid of the denominator:
[tex]\[ 4 \pi r^3 = 100 \][/tex]
Divide both sides by [tex]\( 4 \pi \)[/tex]:
[tex]\[ r^3 = \frac{100}{4 \pi} \][/tex]
Simplify:
[tex]\[ r^3 = \frac{25}{\pi} \][/tex]
To solve for [tex]\( r \)[/tex], take the cube root of both sides:
[tex]\[ r = \left(\frac{25}{\pi}\right)^{\frac{1}{3}} = \frac{5^{2/3}}{\pi^{1/3}} \][/tex]
### Step 4: Verify it is a Minimum
To confirm that this critical point gives a minimum surface area, we need to check the second derivative of [tex]\( S \)[/tex]:
[tex]\[ S''(r) = \frac{d}{dr} \left( 4 \pi r - \frac{100}{r^2} \right) \][/tex]
Differentiate again:
[tex]\[ S''(r) = 4 \pi + \frac{200}{r^3} \][/tex]
Evaluate [tex]\( S''(r) \)[/tex] at [tex]\( r = \frac{5^{2/3}}{\pi^{1/3}} \)[/tex]:
[tex]\[ S''\left(\frac{5^{2/3}}{\pi^{1/3}}\right) = 4 \pi + \frac{200}{\left(\frac{5^{2/3}}{\pi^{1/3}}\right)^3} \][/tex]
Since [tex]\( \left(\frac{5^{2/3}}{\pi^{1/3}}\right)^3 = \frac{25}{\pi} \)[/tex]:
[tex]\[ S''\left(\frac{5^{2/3}}{\pi^{1/3}}\right) = 4 \pi + \frac{200}{\frac{25}{\pi}} = 4 \pi + 8 \pi = 12 \pi \][/tex]
The second derivative [tex]\( S''\left(\frac{5^{2/3}}{\pi^{1/3}}\right) = 12 \pi \)[/tex] is positive, confirming that [tex]\( r = \frac{5^{2/3}}{\pi^{1/3}} \)[/tex] is a minimum.
### Conclusion
The radius [tex]\( r \)[/tex] of the cylindrical container that minimizes the surface area is:
[tex]\[ r = \frac{5^{2/3}}{\pi^{1/3}} \][/tex]
The minimum surface area can be found by substituting this [tex]\( r \)[/tex] back into the original surface area function:
[tex]\[ S\left(\frac{5^{2/3}}{\pi^{1/3}}\right) = 2 \pi \left(\frac{5^{2/3}}{\pi^{1/3}}\right)^2 + \frac{100}{\frac{5^{2/3}}{\pi^{1/3}}} \][/tex]
Simplify each term:
[tex]\[ \left(\frac{5^{2/3}}{\pi^{1/3}}\right)^2 = \frac{5^{4/3}}{\pi^{2/3}} \][/tex]
So,
[tex]\[ 2 \pi \cdot \frac{5^{4/3}}{\pi^{2/3}} = 2 \pi \cdot \frac{5^{4/3}}{\pi^{2/3}} = 2 \cdot 5^{4/3} \pi^{1/3} \][/tex]
Similarly,
[tex]\[ \frac{100}{\frac{5^{2/3}}{\pi^{1/3}}} = 100 \cdot \frac{\pi^{1/3}}{5^{2/3}} = 100 \cdot 5^{-2/3} \pi^{1/3} = 20 \cdot 5^{1/3} \pi^{1/3} \][/tex]
Adding these terms together gives the minimum surface area:
[tex]\[ S\left(\frac{5^{2/3}}{\pi^{1/3}}\right) = 2 \cdot 5^{4/3} \pi^{1/3} + 20 \cdot 5^{1/3} \pi^{1/3} = 30 \cdot 5^{1/3} \pi^{1/3} \][/tex]
Thus, the radius that minimizes the surface area is [tex]\( r = \frac{5^{2/3}}{\pi^{1/3}} \)[/tex] and the minimum surface area is [tex]\( 30 \cdot 5^{1/3} \pi^{1/3} \)[/tex] cm[tex]\(^2\)[/tex].
[tex]\[ S = 2 \pi r^2 + \frac{100}{r} \][/tex]
we need to find the radius [tex]\( r \)[/tex] that minimizes this function. Let's break this down step-by-step:
### Step 1: Define the Surface Area Function
The surface area [tex]\( S \)[/tex] is a function of the radius [tex]\( r \)[/tex]:
[tex]\[ S(r) = 2 \pi r^2 + \frac{100}{r} \][/tex]
### Step 2: Find the First Derivative
To find the critical points where [tex]\( S \)[/tex] might be minimized, we need to find the first derivative of [tex]\( S \)[/tex] with respect to [tex]\( r \)[/tex]:
[tex]\[ S'(r) = \frac{d}{dr} \left( 2 \pi r^2 + \frac{100}{r} \right) \][/tex]
Using the power rule and the quotient rule, we get:
[tex]\[ S'(r) = 4 \pi r - \frac{100}{r^2} \][/tex]
### Step 3: Solve for Critical Points
Set the first derivative equal to zero and solve for [tex]\( r \)[/tex]:
[tex]\[ 4 \pi r - \frac{100}{r^2} = 0 \][/tex]
This can be rearranged to:
[tex]\[ 4 \pi r = \frac{100}{r^2} \][/tex]
Multiply both sides by [tex]\( r^2 \)[/tex] to get rid of the denominator:
[tex]\[ 4 \pi r^3 = 100 \][/tex]
Divide both sides by [tex]\( 4 \pi \)[/tex]:
[tex]\[ r^3 = \frac{100}{4 \pi} \][/tex]
Simplify:
[tex]\[ r^3 = \frac{25}{\pi} \][/tex]
To solve for [tex]\( r \)[/tex], take the cube root of both sides:
[tex]\[ r = \left(\frac{25}{\pi}\right)^{\frac{1}{3}} = \frac{5^{2/3}}{\pi^{1/3}} \][/tex]
### Step 4: Verify it is a Minimum
To confirm that this critical point gives a minimum surface area, we need to check the second derivative of [tex]\( S \)[/tex]:
[tex]\[ S''(r) = \frac{d}{dr} \left( 4 \pi r - \frac{100}{r^2} \right) \][/tex]
Differentiate again:
[tex]\[ S''(r) = 4 \pi + \frac{200}{r^3} \][/tex]
Evaluate [tex]\( S''(r) \)[/tex] at [tex]\( r = \frac{5^{2/3}}{\pi^{1/3}} \)[/tex]:
[tex]\[ S''\left(\frac{5^{2/3}}{\pi^{1/3}}\right) = 4 \pi + \frac{200}{\left(\frac{5^{2/3}}{\pi^{1/3}}\right)^3} \][/tex]
Since [tex]\( \left(\frac{5^{2/3}}{\pi^{1/3}}\right)^3 = \frac{25}{\pi} \)[/tex]:
[tex]\[ S''\left(\frac{5^{2/3}}{\pi^{1/3}}\right) = 4 \pi + \frac{200}{\frac{25}{\pi}} = 4 \pi + 8 \pi = 12 \pi \][/tex]
The second derivative [tex]\( S''\left(\frac{5^{2/3}}{\pi^{1/3}}\right) = 12 \pi \)[/tex] is positive, confirming that [tex]\( r = \frac{5^{2/3}}{\pi^{1/3}} \)[/tex] is a minimum.
### Conclusion
The radius [tex]\( r \)[/tex] of the cylindrical container that minimizes the surface area is:
[tex]\[ r = \frac{5^{2/3}}{\pi^{1/3}} \][/tex]
The minimum surface area can be found by substituting this [tex]\( r \)[/tex] back into the original surface area function:
[tex]\[ S\left(\frac{5^{2/3}}{\pi^{1/3}}\right) = 2 \pi \left(\frac{5^{2/3}}{\pi^{1/3}}\right)^2 + \frac{100}{\frac{5^{2/3}}{\pi^{1/3}}} \][/tex]
Simplify each term:
[tex]\[ \left(\frac{5^{2/3}}{\pi^{1/3}}\right)^2 = \frac{5^{4/3}}{\pi^{2/3}} \][/tex]
So,
[tex]\[ 2 \pi \cdot \frac{5^{4/3}}{\pi^{2/3}} = 2 \pi \cdot \frac{5^{4/3}}{\pi^{2/3}} = 2 \cdot 5^{4/3} \pi^{1/3} \][/tex]
Similarly,
[tex]\[ \frac{100}{\frac{5^{2/3}}{\pi^{1/3}}} = 100 \cdot \frac{\pi^{1/3}}{5^{2/3}} = 100 \cdot 5^{-2/3} \pi^{1/3} = 20 \cdot 5^{1/3} \pi^{1/3} \][/tex]
Adding these terms together gives the minimum surface area:
[tex]\[ S\left(\frac{5^{2/3}}{\pi^{1/3}}\right) = 2 \cdot 5^{4/3} \pi^{1/3} + 20 \cdot 5^{1/3} \pi^{1/3} = 30 \cdot 5^{1/3} \pi^{1/3} \][/tex]
Thus, the radius that minimizes the surface area is [tex]\( r = \frac{5^{2/3}}{\pi^{1/3}} \)[/tex] and the minimum surface area is [tex]\( 30 \cdot 5^{1/3} \pi^{1/3} \)[/tex] cm[tex]\(^2\)[/tex].
Your visit means a lot to us. Don't hesitate to return for more reliable answers to any questions you may have. We appreciate your time. Please revisit us for more reliable answers to any questions you may have. Westonci.ca is your trusted source for answers. Visit us again to find more information on diverse topics.
