Find the maximum value of the function f(x,y)=2x2+3y2−4x−5 on the domain x2+y2≤100.

Question

Find the maximum value of the function f(x,y)=2x2+3y2−4x−5 on the domain x2+y2≤100.

Answer 1

The domain x²+y²=100 -> x²+y²=10² represents the interior and boundary of a circle centered at the origin with a radius of 10.

2x²+3y²-4x-5=f -> 2(x-1)²+3y²=f+7 -> (x-1)²/[(f+7)/2]+y²/[(f+7)/3]=1 represents a horizontal ellipse centered at the point (1,0) with a²=(f+7)/2 (square of the semi-major axis length) and b²=(f+7)/3 (square of the semi-minor axis length). The value of f(x,y) is constant for all points (x,y) on this ellipse.

Increasing the value of f has the effect of increasing the lengths of both the semi-major and semi-minor axes by the same factor. Notice that when f=-7 the lengths of these two axes is 0 and the ellipse degenerates to a point at (1,0). Values of f below 7 produce negative lengths so those values can be ignored. Increasing the value of f from -7 initially creates an ellipse completely enclosed by the given circular region. At some value of f the rightmost vertex of the ellipse will touch the circle at the point (10,0). Past that value the ellipse will touch the circle at 2 points, both of which will have the same x-value. Next we reach a value of f at which the ellipse will touch the circle at the point (-10,0) (that's 3 points of contact in total now). Past this value of f the ellipse will touch the circle at 4 points, 2 of which share one x-value and the other 2 of which share a different x-value. Finally at some value of f the ellipse will touch the circle at 2 points only (both of which share the same x-value), and it is at this value of f that the ellipse is tangent to the circle. Past this value the ellipse will contain the entire circle (and thus the circle will not contain points (x,y) with f(x,y) equal to those values of f). Our goal is to find the value of f where the ellipse is tangent to the circle, and we are aided in this task by the observation that any further increase in f past this value will result in 0 points of contact between the ellipse and the circle.

Those points of contact can be represented by the system of equations representing the circle and the ellipse:

x²+y²=100

2(x-1)²+3y²=f+7

Since y²=100-x² we can plug in that expression for y² in the equation for the ellipse and solve for x:

2(x-1)²+3y²=f+7 -> 2(x-1)²+3(100-x²)=f+7 -> 2(x²-2x+1)+300-3x²=f+7

-> 2x²-4x+2+300-3x²=f+7 -> -x²-4x+295-f=0 -> x²+4x+f-295=0

Using the quadratic formula we obtain

-> x=[-4±√[16-4(f-295)]]/2 -> x=(-4±√Δ)/2

Notice that if the discriminant Δ=16-4(f-295) in this formula is less than 0 we obtain 0 solutions, which corresponds to 0 points of contact between the ellipse and the circle. Let us solve for the values of f which yield this situation:

Δ<0 -> 16-4(f-295)<0 -> 4-(f-295)<0 -> f>299

Therefore, for values of f above 299 we obtain 0 points of contact between the ellipse and the circle. Since this range of values of f includes arbitrarily large values of f, and since such values correspond to the case in which the ellipse entirely encloses the circle without touching it, it follows that at f=299 the ellipse is precisely small enough (but no smaller) to be tangent to the circle at 2 points sharing the same x value. Indeed, the case f=299 corresponds to Δ=0, and thus to the case in which there is only 1 solution for x. It follows that this value of x is:

x=(-4±√0)/2=-2

Since y=±√(100-x²) it follows that:

y=±√[100-(-2)²]=±√(100-4)=±√96=±4√6

Therefore, the values of (x,y) at which f(x,y) is maximized within the given domain are (-2,±4√6), and the value of f(x,y) at those points is 299.