Question

All rectangles with an area of 64 have a perimeter given by $P(x)=2 x+128 / x,$ where $x$ is the length of one side of the rectangle. Find the absolute minimum value of the perimeter function on the interval $(0, \infty) .$ What are the dimensions of the rectangle with minimum perimeter?

          All rectangles with an area of 64 have a perimeter given by $P(x)=2 x+128 / x,$ where $x$ is the length of one side of the rectangle. Find the absolute minimum value of the perimeter function on the interval $(0, \infty) .$ What are the dimensions of the rectangle with minimum perimeter?
        
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Calculus: Early Transcendentals
Calculus: Early Transcendentals
James Stewart 8th Edition
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All rectangles with an area of 64 have a perimeter given by $P(x)=2 x+128 / x,$ where $x$ is the length of one side of the rectangle. Find the absolute minimum value of the perimeter function on the interval $(0, \infty) .$ What are the dimensions of the rectangle with minimum perimeter?
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Transcript

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0:00 Hi there.
00:01 So for this problem, we are given the perimeter for this, that is p of adds, is equal to two times x, and this plus 128 divided by x, okay? so with that said, the question for this is to find the absolute minimum value of the perimeter function on the interval between zero and infinity.
00:25 And the other question is, what are the dimensions of the rectangle with the minimum perimeter? with that said, first of all, we're going to take the derivative of the perimeter with respect to x.
00:39 So that will be then in this case to minus 128 divided by x square.
00:46 Then we need to, since we are finding, since we want to find a critical point for this, we are going to set this equal to zero.
00:53 Then we will have 2 is equal to 128 divided by adds a square.
00:59 So solving 4 ads in here, we will obtain that adds is equal to 8.
01:04 So this is a critical point.
01:06 So in order to determine if this point give us a maximum or a minimum, we need to take the second derivative of the function with respect to adds of the perimeter function and then evaluate that at 0.
01:23 So that will be then equal to 2506 divided by x to the 3.
01:32 So let's evaluate that now at the value that we obtained from before, which is 8.
01:38 Now that will be 256 divided by 8 to the 3.
01:43 So the value that we obtained from this is equal to 1 divided by 2, which is a value that is greater than 1.
01:50 So since the second derivative is greater than 1, we have a relative minimum at x equals to 8...
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