HW 7 Solutions

HW 7 Solutions - Math E-21a Fall 2009 HW#7 Solutions...

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1 Math E-21a – Fall 2009 – HW #7 Solutions Section 11.7 :

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5 50. Since the plane passing through (1, 2, 3) must cut out a tetrahedron in the 1st octant, its intercepts must all be positive – call them a , b , and c . The volume of this tetrahedron will be 1 6 Va b c . The best form of the equation for this plane is the intercept form, i.e. 1 xyz abc  . [It’s easy to check that this is correct by setting any two of the coordinates equal to zero and solving for the other.] Because (1, 2, 3) must lie on this plane, we get the constraint 123 1  . We can then solve the problem by (a) solving for one of the unknowns in the constraint and substituting this into the volume expression to get a function of two variables; or (b) using the Method of Lagrange Multipliers. Let’s do both. Method (a) : It’s slightly easier to use implicit differentiation in this problem. We have 1 6 b c where we can think of c as a function of a and b , so V can then be viewed as a function of a and b . Differentiating V , we get: 11

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This note was uploaded on 02/10/2011 for the course MATH E-21a taught by Professor - during the Fall '09 term at Harvard.

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HW 7 Solutions - Math E-21a Fall 2009 HW#7 Solutions...

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