Differential Calculus (2017 edition)
- Optimization: sum of squares
- Optimization: box volume (Part 1)
- Optimization: box volume (Part 2)
- Optimization: profit
- Optimization: cost of materials
- Optimization: area of triangle & square (Part 1)
- Optimization: area of triangle & square (Part 2)
- Optimization problem: extreme normaline to y=x²
Sal constructs an equilateral triangle & a square whose bases are 100m together, such that their area is the smallest possible. Created by Sal Khan.
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- at 1.10 why is derivative of sqrt3*x^2/36 is equal to sqrt3*x/18?
why isn`t quotient rule for derivative applied in here?(23 votes)
- The quotient rule is not applicable here because the denominator doesn't involve x. The only term involving x we have is x² ("up top" in the numerator), its coefficient being √3/36.(21 votes)
- Given the exact same problem, how would i solve which it for the maximum area?
I mean when you derivate it gives a minimum point.(6 votes)
- Unless you impose a closed interval of allowed lengths, you don't have a maximum area. In this case, there is a physical measurement the requires that 0 ≤ x ≤ 100
This is a closed interval, so what we do is check the value of the original function, not the derivative, at the two endpoints. The absolute max will be whichever of these gives us the greater area.
In this case, if x = 0 it means all of your material went into building the rectangle, thus you have no triangle. If x=100, then all of the material went into building the triangle and you have no rectangle.
If you check the two endpoints, it turns out you have an absolute max at x=0, that is if you only built a rectangle.
If you specify that you cannot use all of the material to build one structure, that you must have both, then you have an open interval. In that case there is no maximum and no way to maximize the enclosed area.(9 votes)
- At5:30which side of the cut string would you use for which shape? Or would it not matter?(6 votes)
- x is equal to the amount of string used in the triangle, and 100 minus x (the remainder of the string) is used to make the square! Yay math!(8 votes)
- for #1, how should I explain how or why at x=-1 and at x=3 they're absolute max and mins.
Same for #2: For the inflection point at x=1, should I explain by saying that the slope changes signs?(5 votes)
- Since you had a defined, closed interval, the largest value is the absolute max FOR THAT INTERVAL. Similar for the min.
You can show that x=-1 is an absolute max because it is the only local max, the function is continuous, and there are no other extrema on the given interval. Thus, it must be the largest value of f(x) over the interval. In other words, since f(x) is increasing over -3<x<-1, then f(-1) must be the largest value of f(x) over -3<x<-1; and, since f(x) is decreasing over -1<x<3, then f(-1) must be the largest value of f(x) over that interval. Thus, f(-1) must be the absolute max over the given interval.
You can show that f(3) is the absolute min on the given interval because the function is continuous, reaches a max along the interval, has no other local extrema, and f(x) is decreasing as x increases for the interval -1 < x ≤ 3. Thus, the absolute minimum over the given interval must be either f(-3) or f(3), and f(3) is less than f(-3).
You should prove that f(1) is an inflection point with the Second Derivative test:
f''(1) = 0, a necessary condition for an inflection point.
f''(1-ε) is positive AND f''(1+ε) is negative, a sufficient condition for an inflection point. (Note: ε means an arbitrarily small number).
Therefore, f(1) fully meets all the requirements of the Second Derivative Test, thus it must be an inflection point.(4 votes)
- This is a question about some of the optimization exercises; they say "The trick here is to recognize that the x value that will maximize the value of S(x) is the same as the x value that maximizes A(x)" in the hints, when S(x) = (A(x))^2. How does that work?
An example is when A(x) = 4xsqrt(r^2-x^2) and S(x) = 16x^2(r^2−x^2).
EDIT: I mean honestly none of these exercises were taught in the videos at all(4 votes)
- The input that will maximize a positive expression will also maximize its square, as the squaring function is a strictly increasing function (on the domain of non-negative real numbers). Sometimes you have to think a bit out of the box to apply these tools to certain problems.(4 votes)
- At3:09, Sal finalizes our answer as a number with a decimal in the numerator and fractions in the denominator.
To express our answer in simplest terms, do we need to combine the two fractions in the denominator?
Also, is it the custom in mathematics when simplifying fractions to give our final answer in either decimals or fractions, but not both, or is it "permissible" to leave our answer in its current form.(4 votes)
- Since Sal's goal is to find a decimal approximation for x, it is fine to leave his work in the form it's in at3:09. If this was to be his final answer, then yes it is conventional to rationalize denominators, not leave fractions with decimal parts, and simplify fractions using a common denominator. The advantage of leaving it the way he does is that the 'fraction' he creates is still an exact value for x. Since he uses a calculator to compute the final answer, no 'clean-up' was necessary.(1 vote)
- If the question had been to maximize combined areas. The answer can't be achieved by finding critical points since the only existing one is a minimum. Does this always mean that there is no way to maximize both areas and therefore one of the figures will have a higher Area/perimeter ratio and therefore all the 100 meter cord should be use to create that figure? I have done the ratio Area/Perimeter of both figures. Being Asq= Area of square; Atr= Area of triangle; Psq= Perimeter of square; Ptr= Perimeter of triangle and X= size of the side of square. These ratios (Asq/Psq = 1/4 X and Atr/Ptr~1/5X) tell me that the solution to optimize areas is to only do a square with all the 100 meter cord. Is there any better way to get to this solution using derivatives?(2 votes)
- It would be the endpoints, not the critical points. Remember the maxes and mins are either at endpoints or critical points!(2 votes)
- At1:30Sal differentiates (100-x)^2/16 as (100-x)/8(-1), wouldnt you just use the chainrule or even the quotient rule? I thought the derivative would've been ((2(100-x)(16)-(100-x)^2))/(16^2) ?(1 vote)
- At1:12, why is the derivative of sqrt(3)x^2/36 = root 3x/18? I tried using chain rule and quotient rule on that term and I ended up getting 4 root 3x as my final answer. Was there an easier way to do that, or a correct way rather?(0 votes)
- No need for the chain rule or the quotient rule here. All you have is x^2 times a constant (which happens to be sqrt(3)/36). The derivative of x^2 is 2x, and you multiply that by the constant to get the result indicated.(6 votes)
- Am I right in thinking that the use of the 2nd derivative test is not necessary in order to verify a min/max value at a critical point?
Using the example in the video - we know that the CP = 0 at x = 56.5. I know this final calculation is done at the end but you can easily do it when you are establishing the CP=0 value.
Then by taking 2 values- one slightly above and one below (eg 55.5 and 57.5) and plugging them back into the 1st derivative equation you get -0.22 and +0.22 - meaning the curve is transitioning from negative to zero to positive.
This confirms an upward curve and thus x =56.5 as the minimum value.
Am I correct in my thinking?(2 votes)
- You are correct. However, this is assuming that there is not a sudden turn in the graph between the maximum that you are confirming, and the point nearby that you are using to check the slope.
The beauty of the second derivative check is that you know that you are actually checking the point you care about, and there can be no changes nearby that would effect your results.(2 votes)
Where we left off in the last video, we had come up with an expression as a function of x of our combined area based on where we make the cut. And now we just need to figure out where this hits a minimum value. And to do that, we just have to take the derivative of this business, figure out where our derivative is either undefined or 0, and then just make sure that that is a minimum value, and then we'll be all set. So let me rewrite this. So our combined area as a function of x, let me just rewrite this so it's a little bit easier to take the derivative. So this is going to be the square root of 3 times x squared over-- let's see, this is 4 times 9. This is x squared over 9. So this is going to be 4 times 9 is 36. And then over here in blue, this is going to be plus 100 minus x squared over 16. Now let's take the derivative of this. So A prime, the derivative of our combined area as a function of x, is going to be equal to-- well, the derivative of this with respect to x is just going to be square root of 3x over 18. The derivative of this with respect to x, well, it's the derivative of something squared over 16 with respect to that something. So that's going to be that something to the first power times 2/16, which is just over 8. And then times-- we're just doing the chain rule-- times the derivative of the something with respect to x. The derivative of 100 minus x with respect to x is just negative 1, so times negative 1. So we'll multiply negative 1 right over here. And so we can rewrite all of that as-- this is going to be equal to the square root of 3/18 x plus-- let's see, I could write this as positive x/8. So I could write this as 1/8 x, right? Because a negative 1 times a negative x is positive x/8. And then minus 100/8, which is negative 12.5-- minus 12.5. And we want to figure out an x that minimizes this area. So this derivative right over here is defined for any x. So we're not going to get our critical point by figuring out where the derivative is undefined. But we might get a critical point by setting this derivative equal to 0 to figure out what x-values make our derivative 0. When do we have a 0 slope for our original function? And then we just have to verify that this is going to be a minimum point if we can find an x that makes this thing equal to 0. So let's try to solve for x. So if we add 12.5 to both sides, we get 12.5 is equal to-- if you add the x terms, you get square root of 3/18 plus 1/8 x. To solve for x, divide both sides by this business. You get x is equal to 12.5 over square root of 3 over 18 plus 1/8. And we are done. At x equals this, our derivative is equal to 0. I shouldn't say we're done yet. We don't know whether this is a minimum point. In order to figure out whether this is a minimum point we have to figure out whether our function is concave upward or concave downward when x is equal to this business. And to figure that out, let's take the second derivative here. So let me rewrite the second derivative of all of this business. The second derivative, well, this was the same function as this right over here. So let me rewrite it. So A prime, the derivative of our combined area, was equal to the square root of 3/18 x plus 1/8 x minus 12.5. The second derivative is going to be square root of 3/18 plus 1/8. So this thing right over here is greater than 0, which means we're concave upward for all x's. Concave upwards, which means for all x's. We're kind of doing a situation like this. So if we find an x where the slope is 0, it's going to be over an interval where it's concave upwards. This is concave upwards for all x. So we're going to be at a minimum point. The slope is 0 right over here. This right over here will be a minimum point. So once again, this is going to be a minimum point. Now, if we actually had a 100-meter wire, this expression isn't too valuable. We'd want to get a pretty close approximation in terms of where to actually make the cut, an actual decimal number. So let's use a calculator to get that. So we have 12.5 divided by square root of 3 divided by 18 plus 1 divided by 8 gives us-- and now we deserve our drum roll. This is 56.5. So this is approximately equal to 56.5 meters. So you make this cut roughly 56.5-- I'll write roughly-- 56.5 meters from the left-hand side. And you will minimize the combined area of both of these figures.