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# Optimization Using the First Derivative Test - Problem 3

###### 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.

When given a line that passes through a point (8,5), intersecting the x-axis at (a,0) and y-axis at (0,b), it is possible to find the minimum area of the triangle formed between (a,0), (0,b), and the origin (0,0). Recall that the area of a triangle is A=½bh, where b is the base and h is the height. So, in this case, the base is a and the height is b so A=½ab.

The slope of the line passing through (8,5) is b/a (since the slope of a line is simply the change in y divided by the change in x). The slope of the line can also be found by finding the slope between (0,b) and (8,5), which can be used to find the slope between (8,5) and (a,0). Setting these slopes equal to one another, you can find a relationship between a and b. Using this relationship, can plug it into the equation for A=½ab. This will create a quadratic equation that you can derive and find the derivative of in order to find the minimum value.

Let’s do a slightly tougher problem. A line through point 8,5 intersects the x axis at point a with coordinates a,0 and the y axis at point b with coordinates 0,b forming triangle AOB. Now O in this case is the origin find the minimum area of triangle AOB if a and b lower case, are positive.

So we want to find the minimum area of this triangle here. AOB as long as we keep these values A and B positive. So you can see I’ve got a triangle here I just draw a random triangle formed by this line. First of all let’s observe that the area of the triangle is going to be ½ base times height. What’s the base in this case?

This length is the base and that’s just a little a. And this is the height and that’s lower case b. So this is ½ a times b. Now the problem with this is you’ve got two variables. And so I have to try and find a way to express area in terms of just one of these two variables. So I need a constraint.

Now the problem is that the constraint isn’t to be found in the problem it’s in the picture somewhere. It’s this line something to do with this line. Let’s observe that the line has a slope it’s a fixed slope. And we can calculate that slope. It’s the change in y over the change in x.

So one way to calculate it would be to take these two points b and a and do the change in y over the change in x. So 0 minus b could be the change in y, a minus 0 will be then change in x. But another way to calculate the slope will be to use these two points 85 and 0b. 5 minus b it’s the same slope so they are equal and 8 minus 0.

Now let’s see what we got here. Minus b over a, 5 minus b over 8. So this is an equation that relates a and b. And somehow, I have to solve this and use it to substitute for one of the two variables in this equation. So let’s see how we do that. Let’s try cross multiplying here. So actually even better idea why don’t we take the reciprocal of both sides?

I’ll take the reciprocal of both sides now and that will give me –a over b equals 8 over 5 minus b. That’s much better because this way now I can solve for a really easily. So I’ll take this up here I’m going to multiply both sides of this equation by negative b. That will give me. That will solve this equation for a.

Minus a over b equals 8 over 5 minus 5. So multiply both sides by –b. And you see that this is going to be a,and over here we are going to get minus 8b over 5 minus b. So now I’ve solved for a keep in mind that the area is ½ a b. So I’m going to pop this in. My area function becomes ½ this times b minus 8b over 5 minus b times b. And when I multiply that through I get area equals. The one half and the 8 cancel leaving a 4, and I’ll get a b² on top. So minus 4 b² over 5 minus b. That’s my area then let me stop a minute because it’s really important. If you are going to use the first derivative test to define the interval of this thing is defined over.

Let’s look back at our picture again. My variable is b. So clearly I want b to be positive but there’s more of a restriction than that. B has a b definitely has a minimum value. If you think about it I really wouldn’t want b to be any less than 5.

If b is less than 5 my line is going to go like this and it’s not actually going to enclose a triangle at all. If b equals 5 my line will be horizontal and it will enclose a giant rectangle that it’s infinitely long. But if b Is bigger than 5 I’ll get a triangle I will actually get an x intercept. And I’ll have a complete triangle with a finer area.

So I need b to be bigger than 5 I need this point to be above this point. So that’s my domain b is bigger than 5. Very important I need to take the derivative. So let’s call that area prime. And let’s note that this is a quotient it’s going to need the quotient rule.

So that’s going to be 5 minus b. Low d high that's the derivative of the top minus 8b minus the top, times the derivative of the bottom that’s -1. Over the square of the bottom, 5 minus b². Now let’s see what this gives us. We have an 8b² and we have plus 4b².

Sorry minus 4b² so 8b² minus 4b² is 4b². And then we have this value here minus 40b. All that over 5 minus b quantity squared. I should factor this just a little bit more. Let’s see and I’ll put that here. Area prime equals and I can take a 4b out of this and that will leave b minus 10.

4b(b minus 10) over 5 minus b². Let’s look for critical points first of all let’s observe that 5 is not a critical point. Even though the derivative is undefined at 5, 5 is not in the domain of my area function. So it doesn’t count as a critical point but I will put for 5 undefined.

B equals 10 that is a critical point. It's greater than 5 so I want to take a look at that. That’s where the derivative is 0, at b equals 10. 0 again not in the domain. Although 0 does make the derivative. 0 it’s not in my domain. So I don’t need to consider it.

Let’s test some points between 5 and 10 let’s just try 6 for example. I’ll have 5 minus 6 which is -1 but squared is 1. So positive in the denominator 6 minus 10, -4. This will be positive 4 times 6, 24 times -4 that will leave negative on top, positive on bottom that will be negative.

And then to the right of 10, let’s try 11. 4 times 11 positive, 11 minus 10 positive, 5 minus 11, negative but squared will make it positive. This is all positive, so I’m going to get a positive derivative over here. And it looks like we have an absolute minimum at b equals 10.

Remember write your justification. Area prime is less than 0 on the interval from 5 to 10. Area prime is greater than 0 from 5 to infinity. So by the first derivative test for extreme values, area is a minimum at b equals 10. Let’s double check. What's that we are supposed to find?

Find the minimum area of triangle AOB if a and b are positive. So we are actually just supposed to find the minimum area. I just found the b value that’s this value so that’s supposed to be 10. The actual minimum area, I get from this function. So I have to pop the value 10 in there. So a of 10 equals minus 4 times 10² that’s minus 400 over 5 minus 10.

5 minus 10 is -5. So its 400 over 5 that’s 80. So the area is going to be 80 that’s the minimum area of triangle AOB.

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###### Norm Prokup

PhD. in Mathematics, University of Rhode Island

B.S. in Mechanical Engineering, Cornell University

He uses really creative examples for explaining tough concepts and illustrates them perfectly on the whiteboard. It's impossible to get lost during his lessons.

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