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The volume of a right circular cylinder is 128π centimeters cubed.
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Find the height and base radius of the cylinder such that its surface area is as small as possible.
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Letβs start with what we know.
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We know that the volume of a right circular cylinder equals π times the radius squared times the height.
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ππ squared is the area of its circular base.
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And we multiply that by the height of the cylinder.
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For our particular cylinder, the volume equals 128π centimeters cubed.
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And the surface area of a right circular cylinder can be found by multiplying two times ππ squared plus two ππ times the height.
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We want to optimize the surface area, under the constraints or the conditions that the volume must equal 128π centimeters cubed.
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To do that, we wanna take the derivative of the surface area formula.
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However, right now, our surface area has two variables in the formula, the radius and the height.
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And we would prefer to just have one variable.
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We can use our constraints, the volume and the formula for volume, to rewrite the height in terms of the radius.
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To do that, weβll divide both sides of this equation by ππ squared.
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On the right side, weβre left with the height.
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On the left, the π in the numerator and the π in the denominator cancel out.
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So weβll say that the height must be equal to 128 divided by the radius squared.
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And that means, in place of β, the height, in our surface area formula, we can substitute 128 over π squared.
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At this point, we notice that there is an π in the numerator and an π squared in the denominator.
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And that can simplify so that the formula for the area is equal to two ππ squared plus two π times 128 over π.
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We can pull out the factor two π.
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To optimize, we first take the derivative of the surface area formula, the derivative of the surface area with respect to π.
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And taking the derivative, itβll be easier if we rewrite 128 over π as 128 times π to the negative one power.
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We know that our constant, two π, stays the same.
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The derivative of π squared is two times π to the first power.
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Then weβll take the derivative of 128 times π to the negative one power, which is negative 128 times π to the negative two power.
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Like I said earlier, optimizing means taking the derivative of this formula and then setting it equal to zero.
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Two π times two π minus 128π to the negative two power equals zero.
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We can rewrite negative 128π to the negative two power as 128 over π squared.
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And now we need to solve for π.
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First, we divide both sides of the equation by two π.
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Two π divided by two π cancels out.
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And zero divided by two π equals zero.
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On the left, we need to find a common denominator.
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We can multiply two π by π squared over π squared.
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That would give us two π cubed over π squared minus 128 over π squared, which we can simplify to say two times π cubed minus 128 over π squared equals zero.
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Then multiply both sides of the equation by π squared.
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Two π cubed minus 128 equals zero.
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We add 128 to both sides.
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Two π cubed equals 128.
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Divide both sides of the equation by two.
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π cubed equals 64.
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And then take the cube root of π cubed and the cubed root of 64, and π equals four.
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The optimized radius equals four.
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Our volume was given in centimeters cubed, which tells us that our radius is being measured in centimeters.
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To find the height, weβll use the formula that we have that compares the height to the radius.
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The height equals 128 divided by the radius squared.
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The height would be 128 divided by four squared.
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Thatβs 128 divided by 16, which equals eight.
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And again, weβll be measuring in centimeters.
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Before we leave this problem, we can go back and plug in the radius and the height to check and make sure that we have gotten the volume we intended.
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Volume equals ππ squared times height.
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π times four squared times eight equals 128π, which is what our constraints were.
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To create a surface area that is as small as possible while maintaining a volume of 128π, a right circular cylinder would need a radius of four centimeters and a height of eight centimeters.