Page 350, EXERCISES 4.1, Problems No. 1, 2, 7, 19.
Solutions.
Prob. 1. Taking derivatives against 
on both sides of 
, one has 
. So substituing 
, one gets 
.
Prob. 2. The answer is 
.
Prob. 7. Denote the meeting airport by 
. It is clear that the distances between airplane 
(resp. airplane 
) to 
are functions of the time 
.
The distances functions are 
%3D30-250t%2C%20B(t)%3D40-300t#card=math&code=A%28t%29%3D30-250t%2C%20B%28t%29%3D40-300t&id=xRrVf) respectively.
The distances between the two airplanes is denoted by 
#card=math&code=l%28t%29&id=Myc6F). By the Pythagorean theorem, we have 
%20%3D%20A%5E2(t)%2BB%5E2(t)#card=math&code=l%5E2%28t%29%20%3D%20A%5E2%28t%29%2BB%5E2%28t%29&id=sT19X). Taking derivatives against 
, we have
%20%5Cfrac%7Bd%20l(t)%7D%7Bdt%7D%20%3D%20A(t)%20%5Cfrac%7Bd%20A(t)%7D%7Bd%20t%7D%20%2B%20B(t)%20%5Cfrac%7Bd%20B(t)%7D%7Bdt%7D%0A#card=math&code=l%28t%29%20%5Cfrac%7Bd%20l%28t%29%7D%7Bdt%7D%20%3D%20A%28t%29%20%5Cfrac%7Bd%20A%28t%29%7D%7Bd%20t%7D%20%2B%20B%28t%29%20%5Cfrac%7Bd%20B%28t%29%7D%7Bdt%7D%0A&id=EmO55)
Since the required changing rate is the value of 
%7D%7Bd%20t%7D#card=math&code=%5Cfrac%7Bd%20l%28t%29%7D%7Bd%20t%7D&id=q5EWH) at 
, so substituting 
%3D30%2C%20B(0)%3D40%2C%20l(0)%3D%5Csqrt%7B30%5E2%2B40%5E2%7D%3D50%2C%20%5Cfrac%7Bd%20A(t)%7D%7Bd%20t%7D%5Cvert%7Bt%3D0%7D%3D-250%2C%20%5Cfrac%7Bd%20B(t)%7D%7Bd%20t%7D%5Cvert%7Bt%3D0%7D%3D-300#card=math&code=t%3D0%2C%20A%280%29%3D30%2C%20B%280%29%3D40%2C%20l%280%29%3D%5Csqrt%7B30%5E2%2B40%5E2%7D%3D50%2C%20%5Cfrac%7Bd%20A%28t%29%7D%7Bd%20t%7D%5Cvert%7Bt%3D0%7D%3D-250%2C%20%5Cfrac%7Bd%20B%28t%29%7D%7Bd%20t%7D%5Cvert%7Bt%3D0%7D%3D-300&id=Qzg6P), we get
%7D%7Bd%20t%7D%5Cvert%7Bt%3D0%7D%20%3D%2030(-250)%2B40(-300)%2C%0A#card=math&code=50%5Cfrac%7Bd%20l%28t%29%7D%7Bd%20t%7D%5Cvert%7Bt%3D0%7D%20%3D%2030%28-250%29%2B40%28-300%29%2C%0A&id=wT0qK)
and hence %7D%7Bd%20t%7D%5Cvert%7Bt%3D0%7D%20%3D%2030(-5)%2B40(-6)%3D-390#card=math&code=%5Cfrac%7Bd%20l%28t%29%7D%7Bd%20t%7D%5Cvert%7Bt%3D0%7D%20%3D%2030%28-5%29%2B40%28-6%29%3D-390&id=T6bdT) mile/h. That is, the airplanes 
and 
are getting closer with the rate 390 mile/h.
Prob. 19. A sphere of radius 
has surface area 
. So 
%7D%7Bd%20t%7D%3D8%5Cpi%20r(t)%20%5Cfrac%7Bd%20r(t)%7D%7Bdt%7D#card=math&code=%5Cfrac%7Bd%20A%28t%29%7D%7Bd%20t%7D%3D8%5Cpi%20r%28t%29%20%5Cfrac%7Bd%20r%28t%29%7D%7Bdt%7D&id=b1clW). Now 
%7D%7Bd%20t%7D%3D-3#card=math&code=%5Cfrac%7Bd%20r%28t%29%7D%7Bd%20t%7D%3D-3&id=dHFK2)m/sec and 
m. So %7D%7Bdt%7D%20%3D%208%5Cpi%2010%20(-3)#card=math&code=%5Cfrac%7Bd%20A%28t%29%7D%7Bdt%7D%20%3D%208%5Cpi%2010%20%28-3%29&id=thSRk) m^2/sec = 
m^2 /sec. That is, the rate the surface area decreases is 
m^2/sec, approximately 753.6 m^2/sec.
Page 451, EXERCISES 4.7, Problems No. 329, 337, 339, 355.
Solutions.
Prob. 329. We are asked to minimized the fuel consumption function 
%3Dav%2Bb%2Fv#card=math&code=g%28v%29%3Dav%2Bb%2Fv&id=glqWE). Taking derivative against 
, one has 
%3Da-b%2Fv%5E2#card=math&code=g%27%28v%29%3Da-b%2Fv%5E2&id=DBoNh). Setting %3D0#card=math&code=g%27%28v%29%3D0&id=TNJyQ) one gets 
. (Note that 
). Since this is the only critical point, this is the minimum for #card=math&code=g%28v%29&id=ltCWd).
Prob. 337, 339. Now 
.
Prob. 337. We need to find out the minimum and maximum of 
. Since 
, we write 
%3Dxy%3Dx(10-x)%3D-x%5E2%2B10x#card=math&code=f%28x%29%3Dxy%3Dx%2810-x%29%3D-x%5E2%2B10x&id=zUUKt). Because 
, so 
. Using the first derivative test, we see that at 
, the function #card=math&code=f%28x%29&id=dhvV3) attains its maximum 
%3D25#card=math&code=f%285%29%3D25&id=wBPlM).
The function %3Dx(10-x)#card=math&code=f%28x%29%3Dx%2810-x%29&id=QbOyk) attains its minimum at 
or 
. The minimum is 
.
Prob. 339. The funciton is 
#card=math&code=y-%20%5Cfrac%7B1%7D%7Bx%7D%20%3D%2010-x-%5Cfrac%7B1%7D%7Bx%7D%3Dg%28x%29&id=ztC4p) (say), with 
. Using the first derivative test, we see that at 
, the function 
#card=math&code=g%28x%29&id=bNMNn) attains its maximum, which is 
%3D8#card=math&code=g%281%29%3D8&id=q3xyo). Since 
%3D-%5Cinfty#card=math&code=%5Clim_%7Bx%20%5Cto%200%2B%7D%20g%28x%29%3D-%5Cinfty&id=DCHKe), the function #card=math&code=g%28x%29&id=MPSk3) has no mimimum over 
#card=math&code=%280%2C%2010%29&id=hlhyL).
Prob. 355. Suppose the rent is 
(unit: dollar/month). Then the number of apartments rented out is 
. Here 
is a non-negative integer. It follows that the profit function is
%3D(800%2B25x-50)(50-x)%3D25(30%2Bx)(50-x)%2C%20x%5Cin%20%5C%7B0%2C%201%2C%202%2C%20%5Cldots%2C%20%5C%7D.%0A#card=math&code=P%28x%29%3D%28800%2B25x-50%29%2850-x%29%3D25%2830%2Bx%29%2850-x%29%2C%20x%5Cin%20%5C%7B0%2C%201%2C%202%2C%20%5Cldots%2C%20%5C%7D.%0A&id=yvQJA)
