Norm Prokup

Cornell University
PhD. in Mathematics

Norm was 4th at the 2004 USA Weightlifting Nationals! He still trains and competes occasionally, despite his busy schedule.

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Optimization Using the First Derivative Test - Problem 2

Norm Prokup
Norm Prokup

Cornell University
PhD. in Mathematics

Norm was 4th at the 2004 USA Weightlifting Nationals! He still trains and competes occasionally, despite his busy schedule.

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Let’s take a look at another example. Determine the constant a for which f(x) equals x² plus a over x. Define for x greater than 0 has an absolute minimum at x equals 5. Find the minimum value of f for this constant.

Here we have a different kind of a problem. We are told where the absolute minimum is going to be and we have to find the value of this constant a, so that the minimum will actually be there. The first thing we always do when we are looking for absolute max or min is take the derivative. Let me rewrite this function as x² plus a times x to the -1.

For the purposes of differentiation I always like to write a over x this way, because I’m going to use the power rule to differentiate. It. So the derivative would be f prime is 2x the derivative of x² plus a times the derivative of x to the minus 1. That’s minus a x to the negative 2.

It’s very important that I put this in the form of a single fraction. This term represents a fraction. Represents a over x². So I need to get this term to have a denominator of x² so I can combine the two fractions.

And 2x would equal 2x³ over x². So that gives me the denominator I need. Now I can combine these two into the single fraction 2x³ minus a over x². Let’s examine this derivative.

What are the critical points? Is x equals 0 a critical point because it doesn’t look like its defined there. It’s actually not because the original function is not defined for x equals 0.So we can disregard that point. The other critical point is going to occur when this numerator equals 0.

So our critical point is going to be when 2x³ equals a. But we need the critical point to happen at x equals 5. So to get the a value we need we need to plug 5 in here. So we need 2 times 5³ to equal a that’s 125 times 2. 250 so we need a to equal 250.

Now that doesn’t guarantee we are going to have a minimum. it is guaranteed to have a critical point at x equals 5. So our function is now x² plus 250 over x. Now let’s look at the derivative to see if we actually do have an absolute minimum.

So I’m going to use the first derivative test here. So here is my derivative and my critical point is at 5. At 5 the derivative is going to be 0 that’s by design and I’ll plug in a number to the left of 5 like let’s say 4. Looking down here keeping in mind I have 2x³ minus 250 over x². I plug in 4, I’m going to have 2 times 64 128 minus 250 that’s negative over a positive it’s going to be negative.

Let’s also keep in mind that this is undefined at 0, and we are only concerning ourselves with positive values. Now let’s try a value to the right of 5 like 6. Going down here 2 times 6³. 6³ is 216 so twice that is going to be 432 minus 250 is definitely positive. And this is going to be positive 6² so it’s positive over positive which is positive.

So we can see that the derivative is negative. Remember that this function is only defined for x greater than 0. On that interval it’s always negative to the left of 5 and always positive to the right. And that justifies that we have an absolute minimum at x equals 5.

But we should write that down. Your teacher is definitely going to want to see some justification that you actually have an absolute minimum. So f prime of x is less the 0, on the interval between 0 and 5 and f prime of x is greater than 0 on the interval from 5 to infinity.

Therefore f has an absolute minimum at x equals 5. Now let’s go back to the problem and make sure that we’ve answered the question. It says determine the constant a for which our function has an absolute minimum at x equals 5 done and that was 250. Find the minimum value of f for this constant. So I still, need to find the actual minimum value.

I need to evaluate f of 5. So that’s going to be 5² plus 250 over 5. That’s 25 plus 250 over 5. That’s 25 plus and the 250 over 5 is 50. So our answer is 75. That’s the absolute minimum value that occurs at x equals 5.

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