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� 25 GEOMETRY OPTIMIZATION 84 25 Geometry Optimization Give the formulas for the radius and length of a closed, half-cylindrical tank (i.e., having a ”D”-like cross-section and flat ends) that will hold a given volume V . Your objective is to make the tank with minimum surface area. What are the specific dimensions if the volume is 100m3? The volume and surface area of the tank are given respectively by: V = πr 2l 2 A = πrl + πr2 +2rl With the volume fixed, we seek the condition where the variation in A is zero, that is, ∂A/∂r = ∂A/∂l =0, subject to the volume constraint. Following the example of Mid-dendorf and Engelmann (1998), we work this out using Lagrange multipliers, even though substitution as in the first problem could also be used. The area itself is the cost function, and we augment it as A = A + λ(V − πr2l/2) A = πrl + πr2 +2rl + λ(V − πr2l/2) Now we have to set the derivatives of A with respect to r and l to zero: ∂A =0 = πl +2πr +2l − λπrl ∂r ∂A =0 = πr +2r − λπr2/2 ∂l These relations along with the constraint V − πr2l/2=0, form three equations in three unknowns. Use the second equation to write λ in terms of r, and then use the third to write l in terms of r, and then substitute both of these into the first equation, to end up with our explicit solution for radius: V (2 + π) �1/3 r = π2 This is exactly what you get if you use substitution to solve the problem, and in that case, we would put this r back into the constraint equation to solve for l (a little messy). But the Lagrange technique gives us more insight than this: if we use the first and second of the three equations above, we can eliminate λ to find that l =2πr/(2 + π). This is a new property of the optimum solution, and we have simply 2π � V (2 + π) �1/3 l = 2+ π π2 If V = 100m3, we find r =3.735m and l =4.564m.MIT OpenCourseWarehttp://ocw.mit.edu 2.017J Design of Electromechanical Robotic Systems Fall 2009 For information about citing these materials or our Terms of Use, visit:


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