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Revenue R=2r' -x and given that increases from to [5 a) (5 pts) Evaluate the actual change in the revenue.6) (5 pts) Evaluate the differential dR...

Question

Revenue R=2r' -x and given that increases from to [5 a) (5 pts) Evaluate the actual change in the revenue.6) (5 pts) Evaluate the differential dR

Revenue R=2r' -x and given that increases from to [5 a) (5 pts) Evaluate the actual change in the revenue. 6) (5 pts) Evaluate the differential dR



Answers

Use differentials to approximate the change in cost, revenue, or profit corresponding to an increase in sales of one unit. For instance, in Exercise 29, approximate the change in cost as $x$ increases from 12 units to 13 units. Then use a graphing utility to graph the function, and use the trace feature to verify your result. Function $\quad$ $x$-Value $C=0.025 x^{2}+8 x+5 \quad x=10$

Let the profit be defined by P which is equal to -0.5 x cube Plus 2500 X -6000. To approximate the change in profit with differentiate P with respect to X. Using power rules who then obtained the derivative dP over dx which is equal to -0.5 times three x squared the derivative of x cubed plus 2500 times won the derivative of X zero, The derivative of 6000. This then simplifies to negative 1.5 X squared. Place 2500. Using differentials. We then write the change in profit delta P. This is approximately the differential P which is this problem is just negative 1.5 X squared Plus 2500 times the differential X Suppose the sales increased from 52 51 units. Then the change in X which is D X. In this case is equal to one. So this we know this is equal to one. So if that is 50, then we have negative 1.5 times 50 squared was 2500 Times one. That's equal to -37 50 Plus 2500, which is equal to -1250. Therefore the profit will decrease By approximately $1,250 after an increase in sales of one unit.

Let the revenue be defined by our which is equal to 50 x minus 1.5 x squared. To approximate the change in revenue. We differentiate our with respect to X. Using power rule. And so we have the derivative of our with respect to X. That's equal to 50 times one. The derivative of X -1.5 times two X. The derivative of X square. This simplifies to 50 three X. Using differentials. We then write the change in revenue delta are this is approximately the differential are which in this problem is just equal to 50 -3 x times differential X. Suppose the sales increased from 15 units to 16 units. Then the changing next, which is the X in this case is equal to one. So this we know this is equal to one. So if access 15 then we have 50 -3 times 15 Times one, that's 50 -45, Which is equal to five. Therefore, The change in revenue after an increase in sales of one unit Is approximately $5.

Let the cost be defined by sea which is equal 2.5 X squared plus four X plus 10. To approximate the change in costs, we differentiate ceo of respect to X. Using power rule, we then obtain the derivative D. C over D X. Which is equal to .05 times derivative of X squared which is two X Plus four times the derivative of X which is one plus zero. The derivative of 10. This simplifies 2.1 x four. Using differentials. We then write the change in cost delta psi. This is approximately the differential see which in this problem is just equal to .1 x was four times a differential. X Suppose the sales increased from 12 units to 13 units. Then the change in X, which is D X in this case is equal to one, so this is equal to one. So if X is 12 then we have 0.1 times 12 plus four Times DX which is one is equal to 1.2 Plus four Or 5.2. Therefore, The change in costs after an increase in sales of one unit Is approximately $5.20.

Question very full were given the revenue in cost function and we need to find the rate off. Change off the revenue costs and profit functions when existent. And the export GTR uh D X over D t is five. So let's differentiate the revenue function both sides with respect of tea. 50 disasters The transition of excess D X over d d 500.5 registers the transition of excess. Where is two x The X over d to using the power change rule. Let's take the X over d t out. So we are left with 50 minus 500.5 times. Two is one. So that would just be X if you replace expired 10 off demonstrations 40 on 40 times five years, 200 So the rate of change of revenue it comes out as 200. Let's do the same procedure for cost. So we have the cost function at a cost function as 10 X plus three. If you divide a free differentiate both sides by with respect to t, we have the sustained D X over D t and the X over D T has already given us five, so this comes out as 50 This is the rate of change off the cost function with respect. Oh, time Onda. Last one is the profit function. We know that profit as revenue minus cost. So the rate of change off profit with respect toe time is the rate of change of revenue with respectful time minus the rate of change of cost. With respect to time which we already found its 200 minus 50 and 200 minus 50 is nothing but 1 15. So this is what is the value off deep your deity?


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