Homework #6 due Mon 28 Oct (Now due Wed 30 Oct) (Solutions) 

All these optimization problems solvable by 4-step process:
(I) write formula for the thing you want to maximize (or minimize)
     as a function of only ONE letter  (call the function F(x))
(II) find F'(x)   by taking derivative
(III) solve for the critical points x so that  F'(x)=0
(also could be at endpoints or where F'(x) undefined, but that does
   not happen in these problems).  Locations of Mins and maxes can 
   only be at critical points.
(IV) evaluate F(x) at the critical points (and perhaps also
nearby points as a sanity check) if necessary to decide
which are maxes and which are mins and to find out how big F(x) is
at those x's.

All these problems had answers in back of book (but without
explanations of why they were right) so these answers could 
have been used by you as a sanity check in your calculations...

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9p186       2
y(t) = -16 t  + 50 t + 5
y'(t) = -32 t + 50
y'(t) = 0 if t = 50/32 = 25/16 = 1.625
so this t is the critical point
                         2
Now y(1.625) = -16(1.625)  + 50 (1.625) + 5 = 44.1


4p201  maximize
                                1/2      1/2    3/2
Area(a) = width*height = (9-a) a    = 9 a    - a

                  -1/2         1/2
Area'(a) = (9/2) a    - (3/2) a
                                         
Area'(a) = 0   if a=3                      1/2 
As can see by multiplying both sides by 2 a    to get
          9 - 3a = 0.

This leads to a  sqrt(3) X 6  rectangle.  


                  2    2     2              2          2
7p201  we have   h  + w  = 60  = 3600  (so h = 3600 - w )
and wish to maximize

            2              2                3
strength = h  w = (3600 - w ) w = 3600 w - w

                         2
strength'(w) = 3600 - 3 w  = 3 (1200 - w^2)

                                1/2               1/2
strength'(w) = 0   if   w = 1200     then h = 2400.



            2    2    2
15p201  diag  = h  + w   (pythagoras)
and area = hw.  
Since area is CONSTANT we have  h = area/w.
So       2    2          2
     diag  = w + (area/w)
Minimize diag is same thing as minimize square of
diag (which is easier) so we get 

         2                     -3                  -2
    (diag )'  =   2w - 2 area w   = 2w ( 1 - area w  ) 
which is 0 if             2    2
                  area = w  = h
i.e. if rectangle is a square, the diagonal is minimized.


17p201                     3
total cost/hour = 675 + p s    where s=peeed in miles/hour and p=const.

                  3
cost of fuel = p s  = $100/hour if spped = 10 mi/hr, so p = 0.1.

so                     3
cost/hour = 675 + 0.1 s .

Now
                                                   3
cost/mile = (cost/hour) (hours/mile) = (675 + 0.1 s )/s

since s = miles/hour and 1/s = hours/mile

so we need to minimize
             -1        2
C(s) =  675 s   + 0.1 s

and to do that we find
              -2
C'(s) = -675 s   + 0.2 s

and  C'(s) = 0 if
               3                    1/3
   -675 = 0.2 s   if  s = [(5)(675)]   = 15 miles/hour.

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