Suppose that
. What is the maximum value of
?
From inspection, we first notice that the equation
will actually give us a circle. Geometrically we are able to graph out the equation. Thus the question is asking us to pick a point on the circle to maximize
. Here lies the main crux of the problem. We are to maximize not
, but
. Knowing that x and y varies, it is already hard enough to find a maximum of the two values added up let alone attaching constants 3 and 4 to them. A use of calculus of variations would be helpful but fortunately we can proceed using algebraic methods.
Let us first complete the square.
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giving us
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We know that this gives us a circle of radius 8 and centre (7,3). Our job is to pick a point of the circle so as to maximize
where x and y are the respective coordinates of the circle.
How about we take it from the
expression. Suppose I let this equal to a variable, that is
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and rearranging, I get
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I can now graph the above equation and attempt to maximize c. With some knowledge of Cartesian formulas, i see that this is a line of slope
and y-intercept
. So, consider geometrically, we need to find the line with slope
that intersects the circle and has the largest value for its y-intercept. The line is as shown below.

The line will be tangent to the curve and the line segment that is the radius to this tangent point has slope
. Here comes the second hurdle. While you may think you already have the answer, there is actually some difficulty solving for x and y. With the usual methods, we will substitute
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into the equation of the circle, but notice two things. There are two unknowns c and x, remember we have yet to find c, and that we will most probably end with a equation with
and
which will be messy to solve.
Instead, with some geometric intuition, notice that I can express the point as
. I am basically taking reference from the centre and using the knowledge of
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where
is the normal to the line and this normal connects the centre to the point. In this case
giving the ratio of the horizontal and vertical components as 3 : 4 and so we now conveniently solve for a.
Substituting
back into the equation of the circle gives me
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So the point that gives the maximum value of
is
or
which yields the maximum value of
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This is certainly not an easy question. Nonetheless, with some geometric inspection, we can work our way to a solution.