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This example, we are going to solve the following optimization problem.

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So we want to find out what is the biggest cylinder inside this sphere.

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So in order to solve that, we need to find out what does it mean by the biggest cylinder?

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We mean the volume of that cylinder.

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So in order to calculate that volume, we can easily draw a figure like this.

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So this Hajj is actually half of the total height of that cylinder.

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OK, and the volume of that half is equal to Hajj multiplied by the area of this circle.

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So the area of the circle is pi r a square and multiply by each will give you the volume of this side

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of the black cylinder and you multiply by two, gives you the whole volume of this cylinder.

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So the objective function will be equal to off is equal to two pi, our square multiplied by H and the

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decision variables are a, R and H.

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So there are some constraints in this example.

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So if you look at this graph, you can easily see that the are small R squared plus Hajari Square is

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equal to our capital, the square.

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So our capital is the radius of the.

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Big sphere on our small is the radius of the base of that cylinder.

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So we do have that constraint and also we know that the small R is between zero and R.

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Capital, so you see here, and H is as well, the same age is between zero and big R, so in order

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to solve that, we need to use the promo package.

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So here you see from Pottermore environment, we import everything and also we may be using the nonpaying

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and these are the comments she's not going to be used.

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So we define a concrete model.

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It can also define as a as an abstract model.

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No worries about it.

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So R is the parameter, initialize it with number one and also edge and are are both variables and the

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arranging between zero and model that are.

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So instead of defining a new constraint, I just put the boundary of the variable inside the variable

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definition and we can initialize it with some numbers between Zero and R and I use the R here.

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So and also we don't need a constraint.

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For example, see a model that sees the constant that the describing the problem for me expression is

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equal to model R-squared plus model square is equal to model.

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That's our capital, the square.

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So as you already noticed, plomo distinguishes between the R capital and small R.

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Yeah.

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And also the objective function is the following expression here.

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I write it down so it will be two pi r a square multiplied by etch.

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And the sense of the optimization is since the Aurora model is a non linear programming, I can use

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the IP of solver and the result is given here.

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So let's run each type separately.

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I run the first stop and then the second tab and you see the number of the number increases after running

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that and then I use this tab.

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So after passing these top.

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The optimization is solved and everything is restored here in my model, and we can print the output,

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so the value is zero point eighty two, which is equal to zero point fifty eight, and the objective

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function is two point forty two.

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And I have used two points for rounding purpose.

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And also, if you want if you don't want to use the round function, you can easily print the whole

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number here.

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OK, so you can easily update any numbers that you see in this example and adapt it to your own values.

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Thank you very much.