Difficult Functions Question from 1996 AHSME #25

Suppose that x^2+y^2=14x+6y+6. What is the maximum value of 3x+4y?

It took me a while to solve it.

A written solution can be read from http://www.gaussianmath.com/functions/1996AHSME25/1996AHSME25.html

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[solo2wolf]

i would have differentiated the circle equation and equated that to the gradient we know we want and thus you have two equations, that and the original and solve by substitution. your method was beautiful though.great stuff as always.

[donylee]

Yup, I see what you mean. Though I must ask how did you get the gradient we know we want.

Probably there's a theorem, but it seems that to maximize '3x + 4y' we need to turn this into a y = mx + c form, compare it with the circle and then equate the derivative of the circle to m.

[ataggake]

As a Maths teacher myself (with students at this level) I found this very interesting. Something to think about - does the introduction of "c" and especially "a" actually help here, or does it confuse?

Once the concept of tangent is understood (and thus we know the gradient of the line from the centre to the solution point), I would tend to go straight to (y-3) = 4/3 (x-7) substituted into the equation of the circle.

Also interesting to compare with the calc. solution.

[donylee]

Hello ataggake,

I was reading your suggestion. The reason for introducing the 'c' and 'a' is to find the function y=y(x) that can be intersected with the circle to get our desired point, the point which maximized our original function 3x+4y.

As for your method, I got to question how did you get the equation (y-3) = 4/3 (x-7) to begin with. You are ALREADY saying that the line from the centre with gradient 4/3 will give the maximum value. I can't see this, must be some theorem.

[YasinIsse]

Why do you make this videos?

[donylee]

To teach mathematics to those with the desire, patience and curiosity to learn.

[iridethewave]

I solved this with a much more efficient method:

represent x and y in terms of cos and sin, you understand from the equation of the circle that each x can be represented as the radius' x distance from 0 + cos of some variable theta, doing the same with y gives

1) 3x + 4y = 24cos(theta) + 32sin(theta) + 33

substituting cos(theta) = sin(theta + pi/2) into (1) we get

2) 24sin(theta + pi/2) + 32sin(theta) + 33 = 3x +4y

through induction we can infer that said expression is maximized

[scott01019]

Cool I never understood how to do those types of problems until watching this video. Awesome job explaining this. Thank you :).

[347670531]

-36 in the third line should be -9

[lamestguyintown]

cant he pronounce three?

[bobthebuilder123451]

@iridethewave Oh, dang, now I know why parametric functions are useful=)

[SeanieG69]

yes lagrange multipliers would work, they can solve most constrained optimisation problems. however for such a simple constraint we can parametrise our constraint curve in terms of a single variable (theta) and apply methods of single-variable calculus, which are generally simpler.

[colplaya7]

wouldn't lagrange multipliers work?

[hacksign23]

dat is duh gut one!

[maths486]

nice solution heissenburg. i've also tried using pure algebra using the discriminant of the quadratic equation

[iridethewave]

i agree heissenburger, that's quite an effective solution. I now feel like a numerical analysis heau.

[Heissenburger]

good idea, but use the R-formula (R=sqrt(24^2 + 32^2) = 40) to get

it into an expression with one theta varying. so 24cos(theta) + 32sin(theta) = 40 sin(theta + alpha)

[alpha = invtan 24/32, but not important] then obviously the max

of this is 40, and overall max is 40 + 33 = 73.