Here. We're told that a sea of X is a cost of producing X units, then see of X over axes, the average cost per unit for part A. We want to show that if the average costs is minimum, then the marginal cost is equal to the average cost. So here we just have to set the derivative of the average cost equal to zero and then solve. So that is the average cost. So when we go to differentiate, we can just use quotient room, get that that IHS X time see prime of X minus sea of X all over X squared and then setting that equal to zero we get that X time. See, Prime of Axe is equal to see a vax. And that means that see, Prime of X is equal to see of ex all over axe. And that's what That's just what we wanted because, see, prime of axe, that's our marginal cost and then see of X over X. That's the average cost. So that's our solution for party. And then when we get to Part B, were given the cost function 16,000 plus 200 acts plus four times X to the three halfs and we want to find for part one we want to find the cost, the average cost and the marginal cost at 1000 units. We know that the cost was given to us and that's just see of 1000 and that is approximately 342,000 491. And so that's our cost. And then we want the average cost. So that's just see of 1000 all over 1000 and thats approximately 342 point for nine. And then next we want the marginal cost and the marginal cost is just see prime of 1000. So when we differentiate our cost function, we get 200 plus three halfs times, for which is six times the square root of 1000 and thats approximately equal to 389 0.74 And that was our marginal cost at 1000 units. So that is the solution for B. We'll call this be one and now the second part, ask us. It's asking us to find the production level that minimizes theatrics cost, So here we just do the same thing. We differentiate the average cost. But to do that, let's first start with finding the average costs. We know that that's just Segovax over axe and see if X was given to us. So that's 16,000 plus 200 acts plus four times X to the three halfs all over acts. And then we can break this up. We get that that 16,000 over X plus 200 plus four times X to the 1/2. And now when we take the derivative of our average costs, we get that that's minus 16,000 over X squared, plus two all over the square root of X. We want to set that equal to zero. So now we have that 16,000 all over. A tune is equal to X square and all over route acts, which means that 8000 is equal to X to the three halfs and then solving for X, we get that X is equal to 400. And that's the solution for part two of being, because we just wanted to find the production level that minimizes that average cost, and we know that that is a minimum for the following reason. So if we take the limit as X approaches zero of this function, we get that that's infinite, that's infinity, and then the limit as X approaches. Infinity of this function is also infinity. So that means that the value, the critical value that we found must be a minimum value. And so now the last part says to find the minimum average cost. So here we just have toe take CF 400 all over 400 because X equals 400 is where the minimum occurs. And if we do that, we get that that is equal to 320 and that completes the problem.