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$2{x}^2 + 3{y}^2 = {a}^2$

What is the maximum value of $3x+2y$ ?

Not knowing calculus and how to graph a ellipse may make this harder. But is there a way to get around calculus? Please determine the answer in terms of $a$.

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Is your question supposed to be: what is the maximum value of $f(x, y)=3x + 2y$ subject to the constraint $2x^2 + 3y^2 = a^2$? –  Shawn O'Hare Jul 25 at 0:25
    
@ShawnO'Hare: I'm almost certain that is the correct form, based on the source text when I hit 'edit'. –  abiessu Jul 25 at 0:26

2 Answers 2

up vote 10 down vote accepted

The curve $2x^2+3y^2=a^2$, for non-zero $a$, is an ellipse in standard position.

Now consider lines $3x+2y=k$, where $k\ge 0$. As $k$ increases, the line moves outward, parallel to itself. The biggest $k$ consistent with meeting the ellipse occurs when the line $3x+2y=k$ is tangent to the ellipse.

Substitute $y=\frac{k-3x}{2}$ in the equation of the ellipse. We get $$2x^2+\frac{3}{4}(k-3x)^2=a^2.$$ There is tangency when this equation has a double root. Expand. We have after some algebra that $$35x^2 -18kx +3k^2-4a^2=0.$$ The discriminant is $0$ when $(18k)^2-(4)(35)(3k^2-4a^2)=0$. This simplifies to $6k^2-35a^2=0$. That yields $k=\sqrt{\frac{35}{6}}\,a$.

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But why does the largest value exist when the line is tangent to the ellipse? What if it is secant to the circle, when there are two solutions? –  user11355 Sep 13 at 18:52
    
You can then move the secant line sort of upward, parallel to itself, which increases the quantity I called $k$, that is $3x+2y$. If the line $2x+3y=k$ meets the ellipse in two points, the $k$ is not as big as it could be. A picture is essential for the visualization. –  André Nicolas Sep 13 at 18:57
    
Yes, if I draw a picture of a ellipse with the center being the origin, because the slope of the line is negative, if you move the secant line upward the point of tangent goes left, which mean of the x value becomes from positive to negative. K value would then be smaller. Am I missing some details –  user11355 Sep 13 at 19:12
    
If you move the secant line "upward" (parallel to itself) the intercepts increase, so the constant term in the equation $3x+2y=k$ increases. To persuade yourself, forget about the ellipse. Draw the line $3x+2y=6$ and $3x+2y=12$. Or use two other constants on the right. The bigger constant gives you a line which is "above" the line with the smaller constant. Constant $0$ is through the origin, increase constant and the line moves up. –  André Nicolas Sep 13 at 19:23
    
As y value increases, x value decreases. How do you determine the point where k is the largest with the point of tangency? –  user11355 Sep 13 at 19:28

Alternatively, using the Cauchy-Scwarz inequality,

$$(3x + 2y)^2 \le \left(\frac{9}{2} + \frac{4}{3}\right)(2x^2 + 3y^2)$$

But $2x^2 + 3y^2 = a^2$, hence,

$$(3x + 2y)^2 \le \frac{35}{6}\cdot a^2$$ $$3x + 2y \le \sqrt{\frac{35}{6}}\cdot a$$

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Very clever! But it seems to require also showing that the strict equality is possible. Is there geometrical interpretation for using dot product related relation here (orthogonality)? –  FooF Jul 25 at 4:31
    
I'm just wonder when were u taught the Cauchy-Schwarz inequalities in high school? –  user11355 Jul 25 at 14:35
    
@user11355 I attended many Mathematics Olympiad courses in high school. There, they introduced this particular inequality together with bunch of other inequalities, such as the Triangle inequality, Minkowski inequality, Young's inequality, and so on. –  Lee Yiyuan Jul 26 at 2:45

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