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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].
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