MATH105 April 2017
• Q1 (a) • Q1 (b) • Q1 (c) • Q1 (d) • Q1 (e) • Q1 (f) • Q1 (g) • Q1 (h) • Q1 (i) • Q1 (j) • Q1 (k) • Q1 (l) • Q1 (m) • Q1 (n) • Q2 (a) • Q2 (b) • Q3 • Q4 (a) • Q4 (b) • Q5 (a) • Q5 (b) • Q5 (c) • Q5 (d) • Q5 (e) • Q6 (a) • Q6 (b) •
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The method of Lagrange multipliers states that a maximum in the interior of the domain must occur at a place where the following equations are satisfied:
We now compute the derivatives:
So we need to solve the system
If we multiply the first equation by 2, we get
Substituting this into the second equation gives
Finally, we multiply both sides by to arrive at the equation
and solve for y to get
We now plug into the third equation:
This simplifies to
which comes out to
From this, it is clear that
Giving a value of
This can be further simplified to
According to the problem statement, this is all we need to do - the problem stated that we do not need to provide justification. We will provide a justification here nonetheless. The family of points satisfying G(x,y) = 0 form a line in the plane. The family of such points with and with are a line segment with endpoints on the y-axis and x-axis. Because the maximum must either occur at a critical point or at an endpoint of the domain, and because f(x,y) = 0 if either x or y is zero, it follows that f achieves its maximum at the critical point (2500, 3750).