00:01
Here we are given two optimization problems.
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In our first problem, consider a rectangle that is bounded below by the x -axis and bounded above by the semicircle y equal to the square root of 25 minus x squared.
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We want to find the dimensions of the rectangle that will maximize its area.
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We'll start by drawing a diagram of our problem.
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So this is a typical cartesian grid and i will sketch the semicircle y equal to the square root of 25 minus x squared.
00:40
This is the equation of a circle of radius 5.
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Within this semicircle and above the x -axis, we can draw a rectangle.
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This rectangle will have a height of y and a length of 2x.
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We want to find the dimensions x, y that will maximize the area of this rectangle.
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Let's first start by quantifying the area.
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So the area of a rectangle will be equal to 2x times y, where the variables x and y are correlated via the equation of a semicircle.
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So let's insert this expression into our area.
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And we'll find that the area of the rectangle written solely as a function of x is equal to 2 times x times the square root of 25 minus x squared, which can also write as 2 times the square root of 25x squared minus x to the power of 4.
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So the goal is to maximize a.
02:41
And how we're going to do this is to find the critical points.
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And we find the critical points by first evaluating the first derivative of our function.
02:51
So let's calculate a prime.
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And this is the reason why i inserted this x variable inside the square root.
03:03
It's going to make utilizing product rule a little bit...
03:09
Not product rule, chain rule a little more straightforward.
03:14
So differentiating a, we obtain 1 over the square root of 25x squared minus x to the power of 4 times the derivative of the interior times the derivative of 25x squared minus 4x to the power of 4, which will give us 50x minus 4x cubed.
03:53
So now we've just found a prime.
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We want to solve for x when a prime of x is equal to 0.
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And this will be the case when 50x minus 4x cubed is equal to 0.
04:18
To solve for this, let's factor a 25x...
04:23
Excuse me, not 25x, a 2x.
04:25
So 2x times 25 minus 2x squared must be equal to 0, which will give us two possible solutions.
04:48
Either x is equal to 0 or x is equal to 5 over the square root of 2.
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And if x is equal to 0, we find that y will be equal to 5 via this relation.
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And if x is equal to 5 over the square root of 2, we find that y is also equal to 5 over the square root of 2.
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So we have two critical points.
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And which of these critical points will maximize our area? evaluating the area, the area of this first critical point is equal to 0.
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So this does not maximize our area.
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And the area of this second critical point will be equal to 25, which will therefore be our local max.
06:17
So this was the solution to our first problem.
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Now we have a second problem.
06:28
So consider an equilateral triangle and a square whose perimeters yield a sum of 10...