Problems when we test for critical values the function might notīe able to be differentiated, or the derivative might be too difficultįound a simple way to determine which variable to use even though it With this is that if the wrong variable is chosen, there could be In this example, and used the resulting equation instead. Substitute in we could have solved for x instead of y *Now, remember earlier, when I said that writing VĪs a function of one variable was the really tricky part? That isīecause sometimes it isn't obvious which variable to solve for and With, the dimensions which yield the greatest volume for our box are aīase of 5.78 inches and a height of 2.88 inches. So, with 100 square inches of material to work Now, we can use this to solve for the height dimension įortunately, we already derived this equation earlier in the example: Out we now know the length of the base which will give the greatest Since -5.78 isn't in our interval, we can throw it So, finding the critical values of our function along this interval: We know that it can't be more than the square root of 100 therefore As such, we know that xĬan't be a negative value, and since our maximum surface area is 100, We are solving for x, the length of a side of the square base of the box. To determine what kind of value we're looking for. Now that we have our function, we need to essentially do the First Derivative Test For this problem, the easiest way is to use x 2 + 4 xy = 100, solve for y, and then substitute it in for the y in V = x 2 y: We need to combine both, solving for V with just one variable. We have theįunctions V = x 2 y, and S = x 2 + 4 xy = 100.

Really tricky we'll get to why at the end of the example. Now that we have our secondary equation, we need to write VĪs a function of only one variable. Maximizing volume using a set surface area: Surface area, since it is the other quantity mentioned - we are Optimized, so it is our primary equation our seconary equation is In this case, volume is the quantity being Which in optimization problems is the quantity which is being related

Now that we have our primary equation, we need to use those variables to create a seconary equation,

Optimization calculus formulas