Discover a world of knowledge at Westonci.ca, where experts and enthusiasts come together to answer your questions. Discover the answers you need from a community of experts ready to help you with their knowledge and experience in various fields. Our platform offers a seamless experience for finding reliable answers from a network of knowledgeable professionals.
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].
We hope you found what you were looking for. Feel free to revisit us for more answers and updated information. We hope you found what you were looking for. Feel free to revisit us for more answers and updated information. Keep exploring Westonci.ca for more insightful answers to your questions. We're here to help.
