Lagrange Multiplier example problem with two constraints
John Travis
Mississippi College
Spring 2012
Consider the ellipse
$$\frac{x^2}{a^2}+\frac{y^2}{b^2}=1$$
that encloses the circle
$$x^2+y^2 = 2x$$.
Determine values of $a$ and $b$ that minimize the area of the ellipse.
First, one must determine a formula for the area inside an ellipse. For an ellipse in the form
$$Ax^2+By^2 + Cxy = 1$$
the area is
$$Area = \frac {2 \pi}{\sqrt{4AB-C^2}}$$
So, for the ellipse here, $A = 1/a^2$, $B = 1/b^2$ and $C = 0$ yields $Area = \pi a b$.
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In the interactive cell below, experiment with various values of a and b so that the ellipse encloses the circle but has smallest possible area.
Click to the left again to hide and once more to show the dynamic interactive window |
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(0, r1) (r4, 0) (0, 0) (0, 0) (r10, 0) (0, 0) (-3/2*sqrt(2), 1/2*sqrt(3)*sqrt(2)) (-3/2*sqrt(2), 1/2*sqrt(3)*sqrt(2)) (3/2*sqrt(2), -1/2*sqrt(3)*sqrt(2)) (3/2*sqrt(2), -1/2*sqrt(3)*sqrt(2)) (3/2*sqrt(2), 1/2*sqrt(3)*sqrt(2)) (3/2*sqrt(2), 1/2*sqrt(3)*sqrt(2)) (-3/2*sqrt(2), -1/2*sqrt(3)*sqrt(2)) (-3/2*sqrt(2), -1/2*sqrt(3)*sqrt(2)) (0, r1) (r4, 0) (0, 0) (0, 0) (r10, 0) (0, 0) (-3/2*sqrt(2), 1/2*sqrt(3)*sqrt(2)) (-3/2*sqrt(2), 1/2*sqrt(3)*sqrt(2)) (3/2*sqrt(2), -1/2*sqrt(3)*sqrt(2)) (3/2*sqrt(2), -1/2*sqrt(3)*sqrt(2)) (3/2*sqrt(2), 1/2*sqrt(3)*sqrt(2)) (3/2*sqrt(2), 1/2*sqrt(3)*sqrt(2)) (-3/2*sqrt(2), -1/2*sqrt(3)*sqrt(2)) (-3/2*sqrt(2), -1/2*sqrt(3)*sqrt(2)) |
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(2.12132034355964, 1.22474487139159) (2.12132034355964, 1.22474487139159) |
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