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A high-tech company purchases a new computing system whose initial value is $ V $. The system will depreciate at the rate $ f = f(t) $ and will accumulate maintenance costs at the rate $ g = g(t) $, where $ t $ is the time measured in months. The company wants to determine the optimal time to replace the system. (a) Let $$ C(t) = \frac{1}{t} \int^t_0 [f(s) + g(s)] \, ds $$Show that the critical numbers of $ C $ occur at the numbers $ t $ where $ C(t) = f(t) + g(t) $.(b) Suppose that $ f(t) = \left\{ \begin{array}{ll} \frac{V}{15} - \frac{V}{450}t & \mbox{if $ 0 < t \le 30 $}\\ 0 & \mbox{if $ t > 30 $} \end{array} \right.$and $ g(t) = \frac{Vt^2}{12,900} $ $ t > 0 $Determine the length of time $ T $ for the total depreciation $ \displaystyle D(t) = \int^t_0 f(s) \, ds $ to equal the initial value $ V $.(c) Determine the absolute minimum of $ C $ on $ (0, T] $.(d) Sketch the graphs of $ C $ and $ f + g $ in the same coordinate system, and verify the result in part (a) in this case.

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a. $C(t)=f(t)+g(t)$b. 30 months c. $C(21.5) \approx 0.0547 V$d. $\mathrm{t}=0$ or $t=\frac{43}{2}=21.5$

12:50

Frank Lin

Calculus 1 / AB

Chapter 5

Integrals

Section 3

The Fundamental Theorem of Calculus

Integration

Missouri State University

Oregon State University

University of Michigan - Ann Arbor

University of Nottingham

Lectures

05:53

In mathematics, an indefinite integral is an integral whose integrand is not known in terms of elementary functions. An indefinite integral is usually encountered when integrating functions that are not elementary functions themselves.

40:35

In mathematics, integration is one of the two main operations of calculus, with its inverse operation, differentiation, being the other. Given a function of a real variable (often called "the integrand"), an antiderivative is a function whose derivative is the given function. The area under a real-valued function of a real variable is the integral of the function, provided it is defined on a closed interval around a given point. It is a basic result of calculus that an antiderivative always exists, and is equal to the original function evaluated at the upper limit of integration.

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A high-tech company purcha…

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We're told a high tech company purchases a new computing system which has an initial value of V. We're told the system depreciates at a rate F equals F. F. T. Is a function of time T. And will accumulate maintenance costs at the rate G equals G of T. Where T. Is the time measured in months. We're told the company wants to determine the optimal time to replace the system. So that's the point of this exercise is defined this optimal time in part A. We are given the function CFT which is defined to be One over t times the integral from 0 to T. Of F. Of S plus G. Of this. Yes. Were asked Should the critical numbers of this expression see occur at the numbers T where C F T equals F. Of T plus G. M. T. Well to find the critical numbers first redefined the derivative. See prime of T. We'll do this by using the product rule. So see prime of T. This is the derivative of one over T. Which is negative, One over t squared times the integral from 0 to T. Of F. Of X plus Gs. Yes plus one over T. Times. And then using the fundamental theorem of calculus, we take the derivative of this integral and by the first part of the fundamental theorem of calculus, this is one of her tee times and this is just F F T plus GFT. Yeah. Now to find the critical numbers, who want to find when c. prime equals zero. So if C prime of T equals zero. Well then it follows after rearranging this expression that one over T squared times the integral from zero to T. Of F five s plus govs. Ds equals on over tee times. Fft plus GFT. And we can multiply both sides by T. And we get that one over T times the integral from zero to T. Of F. F. S plus G. Of this, D. S equals F. Of T plus GFT. Notice that this left hand expression is the same thing as sea of teeth. And so we get that C. Of T equals F. Of T plus Gmt. And this expression is true. This tells us that C F T equals F F. T plus Gmt. When T is a critical number of C. Then in part B were given another expression. Well, our rate of depreciation fft. So we're told that F. F. T is equal to be over 15 minus B. Over 450 times T. 40 greater than zero, less than or equal to 30 And is equal to zero for all T greater than 30. And we're told that the maintenance costs rate GFT is V times T squared over 12,900 40, greater than zero. Were asked to determine the length of time. Big T. For the total depreciation. Dft. The total depreciation D. M. T. Which is the same as The integral from 0 to T. of the rate of depreciation F. Of S. D. S. We have to find the length of time big T. So the total depreciation equals the initial value of the system V. So first, let's put the first piece of art function F. Of S. Into our total depreciation. So we have our total depreciation Gmt is equal to Integral from 0 to T. Of V over 15 -3. Over 450 times S. D. S. And now we'll integrate. So we get the D. F. T. Is equal to these squared, sorry, The over 15 times T uh minus The over 900 times T. Squared. And now we want to solve the equation T. F t equals V. So D of big T is going to be equal to our initial value V. Well, this means that the over 15 times big t minus The over 900 times big t squared equals V. And so we obtained this quadratic equation, which we can solve. It actually reduces to a simpler form. Multiply through by 900 over v. And you should get 900 equals 60 t minus big t squared, Treating a factor as T -30 squared equals zero And therefore big t equals 30. And recall that the unit is in months. And so this is the length of time for the total depreciation to equal the initial value. Then in part C were asked to determine the absolute minimum of the expression. See On the interval from 0 to big teeth, find the absolute minimum on this interval. Let's plug in the expressions F. And G. Into the expression C. Of T. So recall what C. F. T was defined to be, which is one over T times the integral from zero to T. Of Fft. Which 40 less than 30 less than or equal to 30. I should say this is V over 15 minus V over 4 50 S. And then G. Is plus V. S squared Over 12,900 Ds. And now will integrate to obtain the polynomial form of C. So this is equal to one over T. Times view over 15 T minus V over 900 T squared plus V over 38,700 times T. Cute. And this simplifies to view over 15 -3. 900 T Plus V over 38,700 times T squared. Now we already saw in part a that C has critical numbers when C F T. Is equal to F. F T plus GFT. So to find absolute minimum this is going to be at a critical number on this interval. And so we have our left hand side which was CFT. This now becomes FFT plus G of T. On the right hand side is the same V over 15 -1 fee over 900 T Plus V over 38,000 700 he squared. And now we can plug in the expressions for F and G again. So we get be over 15 -3 over 450 T plus the T squared. Over 12,900 Equals v over 15 minus V. Over 900 T Plus the over 38,700 T cubed. Sorry, it's T squared. And so we end up with a quadratic expression and if you want to simplify this, this should factor eventually as tee times to t minus 43 equals zero. And so It follows that are critical numbers are either T equals zero or T equals 43/2. Now because we're only looking at critical numbers in the interval left parentheses zero right bracket. T. We're only considering T equals 43/2. Now. To determine if this is in fact the location of the absolute minimum of seat will test the critical numbers and the end points of the interval. So we have zero which is both an end point and kind of a critical number here you plug it in to see we simply get using our expression above her. See the over 15 on the other hand, plugging in 43/2. Sea of 43/2ves Is be over 15 -3. 900 Times 43/2ves Plus v. 38,700 Time is 43/2ves squared. And this is all an expression in V. Which you can simplify. The first expression was approximately .0667 v. And his second expression see of 43/2 is approximately .054, 7 times the just think you less. And finally we test the right endpoint. So big T. Could be anything from as small as well. zero I guess up to 30 we don't have We do have T. That was 30. Right? So see of 30 Plugging this in this is V over 15-. v. 900 times 30 Plus V over 38,700 times 30 squared. And this is approximately .0566 V. Which is slightly greater than c. 43/2. Therefore, we see the lowest of these. Sea of 43/2ves We saw was approximately .0547 times V. Is the absolute minimum On the interval from zero big teeth. Okay. So in other words, this is the optimal time to replace the system At 43/2 months because this is when the value of C will be the least. Mhm. Okay. Finally, in part the rest to sketch the graphs of C and F plus G and the same coordinate system and verify the result from part a call apart A was about the critical numbers see So we've already found the expressions for C and F plus G. In previous exercises, I'm just going to sketch the graphs very quickly here. The X axis will range from 0 to 30. The y axis from zero to about 10. I should really say T since variable t access by access. Yeah. So a graph C. Of T in red, it starts in the Y axis At the zero v. Over 15. So technically we weren't told exactly what V. Is. But If just for the sake of argument, we say that the equals, then it makes it easier to graph these Soviet for 15 in that case would be a little bit less than seven and it's going to decrease to the point 21.5. And then if you plug in when a 100 To the expression for the absolute minimum, about 5.47- two. And so that's an absolute minimum. So we decrease smoothly until you reach that point and then we start increasing again like this. And so this is CFT. Then I'll also draw F plus G. And blue. This also has a y intercept of zero V. Over 15, And it also passes through the .21.55 .47- two. Since we know that 20.5 is a critical number. Well, just by graphing, we know this is true. Yeah, but on the other hand, F plus G has a minimum much sooner than that, so it looks more like this and this is effortless G. And so we see that the absolute minimum of c occurs at T equals 21.5, which is of course where it intersects with F plus G. And so this confirms part A.

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