Agriculture Paris ^14 created a mathematical model showing plant responses to phosphorus fertili

Agriculture Paris ^14 created a mathematical model showing...

Key Concepts

Chain Rule and Differentiating Radical Functions

The chain rule is a fundamental technique used for differentiating composite functions, particularly when the function includes radical expressions such as square roots. When a square root function is involved, the chain rule facilitates computing the derivative by differentiating the outer square root function and then multiplying it by the derivative of the inner function.

Optimization

Optimization involves finding the maximum or minimum value of a function, which is a central concept in calculus. In many applications, such as maximizing agricultural yield, this involves identifying the input value that leads to the greatest output by analyzing the behavior of the function.

Derivatives

Derivatives measure the rate at which a function changes with respect to changes in its input. In optimization problems, taking the derivative of the yield function with respect to the input variable and setting it equal to zero helps ascertain the critical points where the function could potentially reach a maximum or a minimum.

Critical Points

Critical points occur where the derivative of a function is zero or undefined. In the context of maximizing yield, these points represent candidates for optimal values. After finding these points, further analysis, like the second derivative test, can confirm whether they correspond to a maximum yield.

Transcript

00:01 So if we want to try to figure out what levels of nitrogen and phosphorus are going to maximize our yield equation, we're first going to need to take the derivative of this with respect to x and y, set that equal to zero, get these a and b points, and then just plug it into our second partial derivative test that they give us in the book, and then just see which of these categories it falls into.

00:27 So let's go ahead and find our partials.

00:30 So i'll do the partial with respect to n first.

00:40 So we have n times e to the negative n p.

00:47 Or negative n minus p.

00:52 So we're going to have to use product rule.

00:54 So remember this k and p are constant, so we can pull those out front.

00:57 So it would be k p.

01:00 Or actually let me write this is y sub n.

01:06 And this is all pull that out first of kp and then we have del by del n of n e to the negative p my or negative n minus p so again product rules would be kp times well derivative n is just one to be e to the negative n minus p and then plus n times and the derivative e to the derivative e to the derivative e to the derivative e to the anything is first itself and then we take the derivative on the inside.

01:39 So that is just going to be negative 1.

01:44 And now we can go ahead and factor out this e.

01:51 So it would be k p.

01:54 E to the negative n minus p and then it would be 1 minus n.

02:03 Okay.

02:04 So if we're trying to find what values maximize this, first we're going to set this equal to 0.

02:12 So the only things that are going to be variables in this case is p and one minus n.

02:18 So that's going to imply either p is zero or one minus n is zero.

02:24 So in is going to be one.

02:27 So we have that.

02:29 Now let's go ahead and repeat this for the other derivative.

02:39 So let's take the partial of this with respect.

02:47 To p and actually notice how this is symmetric in the sense of if we were to switch all the ends and p's around we would get the exact same equation so if we come over here and just switch all the variables around we should end up with the exact same derivative as well so this is going to be y sub p and this is going to be equal to so it'd be k in n e to the negative n minus p and then 1 minus p if we just switch the variables from up here and if you just take the derivative like normal you'll see we get the same thing but i will just kind of skip a little bit to make our work a little bit easier so now we're going to go ahead and solve for this so this implies that so let's see either n is zero or 1 minus p is 0 which that gives us p is equal to 1 now if we're trying to find what value to maximize this we can actually go ahead and throw a couple of these out just by looking at this so notice that n cannot be zero nor can p be equal to 0 because if these are equal to 0 we just get 0 for both of those so that is going to imply that n has to be 1 and p has to be equal to 1 because again if we use n is equal 0 p is equal to 0 we have 0 times something so i'll just kind of write that because 0 0 is equal to 0 and that's not going to be a maximum value.

05:09 So now we would just go ahead, take this, plug it into here, and that should actually give us our max.

05:17 So let's see what we get.

05:19 So y11 is going to be, so it would just be k, 1 times 1, e to the negative 1, minus 1, which would be 2, or not 2, just k times e to the negative 2 or you could just have k over e squared so this is where our max is so this is the max yield and that is going to be when n is equal to 1 and p is equal to 1 and i guess we could really just stop here if we wanted because we really don't have anything else to check but just to make sure that this is a maximum.

06:15 We should follow through with these steps over here.

06:19 So again, i mean, if you are just content with what we're doing here, you could just go ahead and say the maximum yield is at n is equal to 1, p is equal to 1.

06:31 But let's actually see if we can confirm that this will be a max.

06:40 So let me come up here.

06:46 So where's the derivative? here it is.

06:48 So this is supposed to be y in.

06:56 So y in and then yp is this here.

07:13 Now if we were to go ahead and bring this down here to see what we need...

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