Let's optimize

by Nuntawatt Vongsavanh

Submitted : Spring 2017

Being that we have a constraint, we must use the given materials efficiently. This ensures that we are building an area with the most with what we are given, plus without finding the largest area we may run into issues. Issues that may arise would be having to purchase more materials, this would be a problem if abiding to strict budgets. First, we start with two functions, one will be the maximization function and the other the constraint. Given those two functions we are able to solve for the constraint for one of the two variables, next we will substitute this into the area and we will have a function of a single variable. Using the beginning constraint function we can solve for ‘x’, substituting what ‘x’ equals into the first maximize function we will have a function of ‘y’.

With the function of ‘y’ we will next take the derivative of that function, next we can set the function equal to zero to solve for ‘y’. This will give us a value of the dimensions of largest area possible for the ‘y’ side of the fence. Solving for ‘y’ we can take that value and find our ‘x’, to do so we will plug in our ‘y’ in our original constraint function when we solved for ‘x’. Once this is done we are given dimensions that will yield largest possible enclosed area. It may seem difficult to understand, but the solution will become clear when explained with visuals in the solution page ahead. 

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