Area of a box is simply the sum of the areas of each

This preview shows page 65 - 67 out of 70 pages.

area of a box is simply the sum of the areas of each of the sides so the constraint is given by2xy+ 2yz+ 2xz= 64=xy+yz+xz= 32.Note that we divide the constraint by 2 to simplify the equation a little. Also, we get the functiong(x, y, z)from this.g(x, y, z) =xy+yz+xz.Here are the four equations that we need to solve.yz=λ(y+z)(fx=λgx),(3.4)xz=λ(x+z)(fy=λgy),(3.5)xy=λ(x+y)(fz=λgz),(3.6)xy+yz+xz= 32(g(x, y, z) = 32).(3.7)Although the equations are nonlinear, there are many ways to solve this system.We will solve it in thefollowing way. Let us multiply equation (3.4) byx,equation (3.5) byyandequation (3.6) byz.xyz=λx(y+z),(3.8)xyz=λy(x+z),(3.9)xyz=λz(x+y).(3.10)Now notice that we can set equations (3.8) and (3.9) equal. Doing this givesλx(y+z)=λy(x+z),λ(xy+xz)-λ(xy+yz)=0,λ(xz-yz)=0=λ= 0orxz=yz.This implies two possibilities. The first,λ= 0,is not possible since if this is the caseequation (3.4) willreduce toyz= 0=y= 0orz= 0.Since we are talking about the dimensions of a box neither of these are possible so we can discountλ= 0.This leaves the second possibility.xz=yz.109(MATH100)notes100-ch3.pdf downloaded by sclaw from at 2014-09-10 05:46:10. Academic use within HKUST only.
3.Partial DerivativesSince we know thatz6= 0(again since we are talking about the dimensions of a box) we can cancel thezfrom both sides. This givesx=y.(3.11)Next, let us set equations (3.9) and (3.10) equal. Doing this givesλy(x+z)=λz(x+y),λ(xy+yz-xz-yz)=0,λ(xy-xz)=0=λ= 0orxy=xz.As already discussed we know thatλ= 0won’t work and so this givesxy=xz.We can also say thatx6= 0since we are dealing with the dimensions of a box so we havey=z.(3.12)Plugging equations (3.11) and (3.12) into equation (3.7)we gety2+y2+y2= 3y2= 32,y=±r323≈ ±3.266.However, we know thatymust be positive since we are talking about the dimensions of a box. Thereforethe only solution that makes physical sense here isx=y=z3.266 cm.This shows that we have a cube here.We should be a little careful here. Since we have obtained only one solution we might be tempted toassume that this is the dimensions that will give the largest volume. The method ofLagrange multiplierswill give a set of points that will either maximize or minimize a given function subject to the constraint.However, when we get a single solution it may be either a maximum or a minimum. To verify that we indeedhave a maximum, as we want, all we need to do is pick any other point that satisfies the constraint andcheck its volume against the volume of the point we got above. If the volume of the point above is largerthan the second point we will know that we indeed have a maximum.To get the second point let us choosey=z= 2plugging these into the constraint gives2x+ 2x+ 4 = 32,x= 7.Checking the volume at the two points givesf(3.266,3.266,3.266)=34.8376,f(7,2,2)=28.

Upload your study docs or become a

Course Hero member to access this document

Upload your study docs or become a

Course Hero member to access this document

End of preview. Want to read all 70 pages?

Upload your study docs or become a

Course Hero member to access this document

Term
Fall
Professor
LEUNGShingYu

  • Left Quote Icon

    Student Picture

  • Left Quote Icon

    Student Picture

  • Left Quote Icon

    Student Picture