How Do You Approach Modeling and Optimization Problems?

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To effectively approach modeling and optimization problems, start by visualizing the scenario with a clear diagram and labeling all known information. Recognize that maximizing or minimizing a value typically requires differentiation, so be prepared to set the derivative equal to zero. Formulate an equation representing the quantity you want to optimize, ensuring it is expressed as a function of a single variable. This structured approach simplifies the problem-solving process and leads to a clearer path to the solution. Understanding these steps is crucial for tackling complex optimization challenges.
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We have just begun this topic and I'm really confused about how to approach questions, is there any trick or guideline for doing so?

Ex: Consider an isosceles right triangle whose hypotenuse is the x-axis and whose vertex is on the y-axis. If the hypotenuse is 2 units long, we'd have x-intercepts of -1 and 1. What is the largest area possible for a rectangle inscribed in the triangle?
 
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The first thing you need to do in this case is draw a picture of the given geometry and label it with known information. Since the goal is to maximize something, a trigger should go off in your head that you'll need to differentiate a function and set it equal to zero. Third, you need to develop an equation of the thing you are trying to optimize (in this case, area of a rectangle) as a function of a single variable (so you'll have something to solve for). That's pretty much it.
 

Thread 'Proving that convexity implies second order derivative being positive'

There are probably loads of proofs of this online, but I do not want to cheat. Here is my attempt: Convexity says that $$f(\lambda a + (1-\lambda)b) \leq \lambda f(a) + (1-\lambda) f(b)$$ $$f(b + \lambda(a-b)) \leq f(b) + \lambda (f(a) - f(b))$$ We know from the intermediate value theorem that there exists a ##c \in (b,a)## such that $$\frac{f(a) - f(b)}{a-b} = f'(c).$$ Hence $$f(b + \lambda(a-b)) \leq f(b) + \lambda (a - b) f'(c))$$ $$\frac{f(b + \lambda(a-b)) - f(b)}{\lambda(a-b)}...
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