2 Responses to “Applied Optimization For Calculus?”

1. yljacktt says:

   October 14, 2009 at 3:56 am

   Okay, xy=122, with x,y lenghts of sides.
   And we minimize 40x+10x+2(10y)=50x+20y.
   So, substitute x=122/y, so 50x+20y=(122*50/y)+20y.
   Now differinate and set to 0.
   So, -(122*50/y^2)+20=0, or 20=(122*50)/y^2.
   So, y=sqrt((122*50)/20) = sqrt305.
   So, lenght of side with bricks is 122/(sqrt305) = 6.9857 approximately.
   And adjacent side is sqrt305 = 17.46 approximayely.
2. bben08 says:

   October 14, 2009 at 7:34 am

   Let’s name x et y the sides of the rectangle (in foot)
   Objective function to minimize total cost “C”:
   C = (x)(40) + (x + 2y)(10) = 50x + 20y
   subject to Constraint equation:
   The total area = A = x y (in foot^2) = 122 ft^2
   —-> y = 122/x
   put back in the objective function:
   C = 50x + 20y = 50x + (2440/x)
   thus C = C(x) function only of x
   to find the optimum derive in term of x
   dC/dx= [ C (x)] ‘ = 50 – (2440/x^2)
   dC/dx = 0 —> 50 – (2440/x^2) = 0
   —–> x^2 = 2440/50 = 48.8
   —-> x = 6.99 ft
   Recall that if the derivative is positive then the function must be increasing and if the derivative is negative then the function must be decreasing. The following number line gives this information
   oo- ————– (x = 6.99) +++++++++ +oo
   So, from this number line we can see that we have the following increasing and decreasing information
   decreasing when x< 6.99
   Increasing when x> 6.99
   x= 6.99 is the minimum
   Finally:
   the other side of the rectangle will be
   y = A/x = 122/6.99 = 17.45 ft
   The brick wall part will cost:
   40 x = (40 )(6.99)= $279.6
   the 3 metal fence parts will cost:
   10 x + 20y = (10)(6.99) + (20)(17.45) = $ 418.9
   for a total cost of : $279.6+$ 418.9 = $698.5

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