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Find the maximum profit and tho number of units that must produced and s0"d R(x) and cost; C(r). of producing x units are order yield iho maximum proli dollars...

Question

Find the maximum profit and tho number of units that must produced and s0"d R(x) and cost; C(r). of producing x units are order yield iho maximum proli dollars. Astuinc that revenueR(x) = 40x - 0.5x? , C(r) = 4x + 20In order t0 yleld Ihe maximum profit of sl units must be produced and sold , (Simplify your answers. Round Io {he nearest cent as nceded )

Find the maximum profit and tho number of units that must produced and s0"d R(x) and cost; C(r). of producing x units are order yield iho maximum proli dollars. Astuinc that revenue R(x) = 40x - 0.5x? , C(r) = 4x + 20 In order t0 yleld Ihe maximum profit of sl units must be produced and sold , (Simplify your answers. Round Io {he nearest cent as nceded )



Answers

Find the maximum profit and the number of units that must be produced and sold in order to yield the maximum profit. Assume that revenue, $R(x),$ and $\cos t, C(x)$ are in dollars. $$R(x)=50 x-0.5 x^{2}, \quad C(x)=4 x+10$$

Okay on this one. Number 24 we have a total cost of total revenue, and then we have a profit definition from the book. That's revenue minus costs. The money coming in, minus the money coming out. Okay, so we're gonna use these to build a profit equation. That profit equation is going to be something that we can use then to maximize it. So let's start off by getting our profit. Pf Exile figured out. So p of X is gonna be equal to R of X, which is 50 X minus 0.5 X squared, minus this whole deal. Let's not mess up our negative signs. Okay, Um, I'm gonna work through it. This over here is gonna become minus 10 X minus three. That's the easy part to mess up. These will remain the same. So let's go ahead and combine like terms. We're gonna have 40 X. This stays the same. And then we're just left with the minus three at the end. All right, So why do we take the derivative of this? What we're trying to figure out when this equation, which probably looks something like a problem. There is a magic point right? It's the highest point. And at that point there's a corresponding X value. Well, this also happens to be the one spot where the slope is equal to zero. And the slope was we found out in this type of equation is equal to the derivative. Okay, lets ppm X. So when that zero meaning, ah, horizontal line. Right. Then we can go ahead and assume that that's going to be the maximum. There is a way, of course, to determine whether it is the maximum over the minimum. But we're not there anything yet, so let's go ahead and take the derivative inside of equal zero. I'm gonna switch colors here, let's say prime of X. Right, So this is going to be the derivative of it. 40. Um, we end up with minus one x, and then this one's three disappears. Okay, so in order for this to equal zero right, 40 minus some mystery number X is gonna have to be the next equals 40. Okay, so that's what the X value. That's the number of units. So we can come up here and say our number of units right is going to be 40 40 units. I always like to put the label on there. Okay, but then we need to figure out what the maximum profit would actually be. So we have to run the numbers. So 40 units. Okay, we Luckily, we have our equation that we're needing to put into right there like this whole thing. So all we have to do is substitute our number of units to figure out our profit. Soapy of 40 right is going to be 40 times 40 minus zero point five. 40 squared, minus three. Okay, let's figure out what that is. Okay? And that ends up. You're going 1797 when you work throughout. Okay, so we go back up here at the top 1700 97 unit. Oh, that's profit. Make sure we get labels. Right Dollars. Excellent. Thank you very much.

Okay. Number 26. Number 26. We have the following information. Total cost, total revenue. We're going to figure a profit equation and then maximize it. So let's go for it. Profit equation. That is P of X. That's going to equal. They are, which is five x on We're going to subtract this whole equation here. 0.1 X squared plus 1.2 X plus 60. Okay, so we're gonna have to go ahead. Make sure we distribute this across so we can bring our light terms together minus 1.2 X minus 16. I'm gonna move this down a little bit. Yeah. Okay, so we bring our ex terms together, and we're gonna have 3.8 minus this amount minus 60. Okay, so this is our p of X equation. We can go ahead now and find the derivative of B lacks. And that's gonna be 3.8 minus 0.2 X. Linus 60. Okay, So what happens here is we have the derivative, and the derivative is, uh, corresponds to If this were the equation, right, there is some sort of spot right at the top that's the maximum happens in some X. Um, we used P prime of X, right? That's the same as the sloping zero. It's the slope. So if p private X zero we know we have a flat spot, and so we know we have the maximum. Okay, So garden. So that's equal to zero. Oh, you know what? I just did? I just have that minus 60. The end. I did not mean to do that. Bad. Bad, bad. Okay, scratch that. 0.2 x, which also will mean that negative 3.8 is equal to negative 0.2 x. Let's divide both sides by negative 0.2 Negative 0.2 and we'll come up with X equals 1900. And that is again in units, not dollars. That's an X represents. Okay, so 1900 units, so Well, let's go. Put that over here. That's half of our answer. Number of units 1900. Um, we have to find the maximum profit. So we take 1900 we put it through this equation right here. That's our profit equation. right. So we're gonna have to go ahead and take that, and we're gonna substitute in 1900 every time there's an X. So we see if I can at least get that written down before I'd defer to the calculator. Oh, actually, this is PM 1900 isn't it? P of 1900? Okay. Equals 3.8 times 1900. Minus 0.1 times 19 100. Also, those 1900 squared minus 16. So P of 1900 obviously equals 3550. Amazing. Okay, so that's gonna be our dollars. So 3550. Okay. So to reiterate, you're selling 1900 units, and then we are getting $3550 back. Thanks.

Okay. Remember, 25 we have some information here, and we have an R of X and a sea of X. They give these two equations to us and represent the total costumes with revenue. The profit is defined in the book as being the revenue minus the cost, which makes sense. That's how you get profit. And we need to find the maximum profit and in the number of units sold to get that to achieve that. So most of our cow are calculations. They're going to be based on the profit equation. So let's go ahead and write that out in full. This profit p of X, this was good. And put the pieces together here. It's equal to two X, which we get from above, minus this delightfully easy to write bunch of numbers. There we go. Actually, this is pretty realistic. I mean, they wouldn't just be in pictures in real life. Um, go ahead and I'm gonna make sure this is simplified as much as I can. So this negative this minus it has to be carried out through each term. Otherwise you will not get the right answer. Okay? Things were going good. So far we can bring our excess together and say one 0.4 X minus 0.1 X squared. Very small. Minus 30. Okay, now what we need to do is we need to take the derivative this on, then set it to zero. What this does is it allows us when we're visualizing the profit curve. There's this mystery spot here. There's just x value. And at that x value, there is a slope of zero. The derivative is zero. So if we just set this relative to zero, we can solve for X, and it will tell us exactly what this point is. Right. But first we have to find p prime of X. So let's do some differentiation. Um, this just becomes 1.4. This term here becomes two times that 0.0 two x and then that minus 30 is a constant. So it's gone. All right, s so now we're going to solve for X when p of x p prime of excuse me zero. So that's zero. Let's say that 0.2 x equals 1.4. So that means that our X right there equals 70. So that's 70 units, not dollars. Okay, so 70 units with our X. So we got half of our answer, right? So we could go back up here and say the number of units is going to be 70 again. He units. So how do I find my maximum profit? Well, I already got the equation for it right here, remember? Remember this guy from just a few minutes ago, So I'm gonna go ahead and substitute in 70. Every time there's an ex, I'm gonna find it out. So this is pf 70 1.4 time, 70 minus 0.1 times 70 squared minus 30. So what is P of 70 equal? Let's do some calculations. Okay. And by the power invested in me by a calculator, I came up with 19. The PS 70 is 19 so let's go ahead right out of here. The maxim profit is $19. 0, my goodness. Well, you got to make money somehow. Hopefully, we'll get more more methods here. That's a lot of units for 19 bucks, But hey, maybe they're quick to make. All right, thank you very much.

We need to determine the number of units that produce a maximum profit. So A. Is the number in front of X squared B. Is a number in front of X. And then we have C. So to find our maximum, we have to find our vertex which is negative B over two. A. And then we evaluate the function P at negative B over two A. So we have negative 1.68 Divided by two times negative one over 14,000. So we type that in negative one .68 divided by two times negative one divided by 14 1000. There we get 11,760. And so we're going to now evaluate the function P At 11,760. So we get negative one over 14,000 Times 11,760 2.68 times 11760 -4000. So we have -1 divided by 14,000 Times 11,760 Squared Plus 1.68 times 11,760 minus 4000. No me yet. 5,878 yeah 0.4 So this is 5,878.4 So 11,760 units yields a max profit Of $5,878.40.


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