Difference between revisions of "1996 AHSME Problems/Problem 25"
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The only answer choice that can be obtained from this equation is <math>\bf{73}</math> | The only answer choice that can be obtained from this equation is <math>\bf{73}</math> | ||
+ | ==Solution 5 (Lagrange Multipliers)== | ||
+ | First, we move all the non-constant terms of the constraint to one side and assign it to the function <math>g(x,y)</math>: | ||
+ | <cmath>g(x,y)=x^2+y^2-14x-6y.</cmath> | ||
+ | Since we are trying to maximize <math>f(x,y)=3x+4y</math>, we need to solve for <math>x</math> and <math>y</math> in the system | ||
+ | <cmath>\begin{cases}x^2+y^2-14x-6y=6,\\\nabla g(x,y)=\lambda\nabla f(x,y).\end{cases}</cmath> | ||
+ | We have that | ||
+ | <cmath>\begin{align*}\nabla f(x,y)&=\begin{pmatrix}\dfrac{\partial}{\partial x}3x+4y\\\dfrac{\partial}{\partial y}3x+4y\end{pmatrix}\\&=\begin{pmatrix}3\\4\end{pmatrix},\end{align*}</cmath> | ||
+ | and | ||
+ | <cmath>\begin{align*}\nabla g(x,y)&=\begin{pmatrix}\dfrac{\partial}{\partial x}x^2+y^2-14x-6y\\\dfrac{\partial}{\partial y}x^2+y^2-14x-6y\end{pmatrix}\\&=\begin{pmatrix}2x-14\\2y-6\end{pmatrix}\end{align*}.</cmath> | ||
+ | To solve the original system, we can solve for <math>x</math> and <math>y</math> in terms of <math>\lambda</math> using our equations from the gradients, then substitute them into the first equation. We have that <math>x=\frac{3}{2}\lambda+7</math> and <math>y=2\lambda+3</math>. Substituting into the first equation, we have that | ||
+ | <cmath>\begin{align*}\left(\frac{3}{2}\lambda+7\right)^2+(2\lambda+3)^2-14\left(\frac{3}{2}\lambda+7\right)-6(2\lambda+3)&=6\\\frac{25}{4}\lambda^2&=64\\\lambda&=\pm\frac{16}{5}\end{align*}</cmath> | ||
+ | Using the solutions of <math>x</math> and <math>y</math> in terms of <math>\lambda</math> that we found earlier, we have that | ||
+ | <cmath>3x+4y=\frac{25}{2}\lambda+33.</cmath> | ||
+ | Because we are trying to maximize this function, we will use the positive solution for <math>\lambda</math>. Therefore, after substituting, we have that the largest value of <math>g(x,y)</math> that satisfies <math>f(x,y)=6</math> is <math>\boxed{\textbf{(B) }73}</math>. | ||
+ | |||
+ | ~qianqian07 | ||
+ | |||
+ | ==Solution 6 (polar coordinates)== | ||
+ | Completing the square gives the equation of a circle, <math>(x-7)^2+(y-3)^2=64.</math> Seeing that we would like to maximize <math>3x+4y,</math> we parameterize the circle using polar coordinates: | ||
+ | <cmath>\begin{align*} | ||
+ | x&=7+8\cos\theta&y&=3+8\sin\theta. | ||
+ | \end{align*}</cmath> | ||
+ | Then, we have <math>3x+4y=3(7+8\cos\theta)+4(3+8\sin\theta)=33+24\cos\theta+32\sin\theta.</math> Since <math>a\cos\theta+b\sin\theta\le\sqrt{a^2+b^2},</math> the desired answer is <math>33+\sqrt{24^2+32^2}=33+40=\boxed{73}.</math> | ||
+ | - Ultroid999OCPN | ||
==See also== | ==See also== | ||
{{AHSME box|year=1996|num-b=24|num-a=26}} | {{AHSME box|year=1996|num-b=24|num-a=26}} | ||
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[[Category:Circle Problems]] | [[Category:Circle Problems]] | ||
{{MAA Notice}} | {{MAA Notice}} | ||
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Latest revision as of 04:46, 26 January 2023
Contents
Problem
Given that , what is the largest possible value that can have?
Solution 1
Complete the square to get Applying Cauchy-Schwarz directly, Thus our answer is .
Solution 2 (Geometric)
The first equation is a circle, so we find its center and radius by completing the square: , so
So we have a circle centered at with radius , and we want to find the max of .
The set of lines are all parallel, with slope . Increasing shifts the lines up and/or to the right.
We want to shift this line up high enough that it's tangent to the circle, but not so high that it misses the circle altogether. This means will be tangent to the circle.
Imagine that this line hits the circle at point . The slope of the radius connecting the center of the circle, , to tangent point will be , since the radius is perpendicular to the tangent line.
So we have a point, , and a slope of that represents the slope of the radius to the tangent point. Let's start at the point . If we go units up and units right from , we would arrive at a point that's units away. But in reality we want to reach the tangent point, since the radius of the circle is .
Thus, , and we want to travel up and over from the point to reach our maximum. This means the maximum value of occurs at , which is
Plug in those values for and , and you get the maximum value of , which is option .
Solution 2B
Let the tangent point be , and the tangent line's x-intercept be . Consider the horizontal line starting from center of circle (O) meeting the tangent line at K. Now triangle is 3-4-5, , so . Note that the horizontal distance from to the origin is , and the horizontal distance from K to Q is 4, ( of its y coordinate), so the x-intercept is . The value of is 73 at point . Note that this value is constant on the tangent line, so there is no need to calculate the coordinate of . .
Solution 3
Let . Solving for , we get . Substituting into the given equation, we get which simplifies to
This quadratic equation has real roots in if and only if its discriminant is nonnegative, so which simplifies to which can be factored as The largest value of that satisfies this inequality is , which is .
Solution 4 (Using Answer Choice + Calculus)
Implicitly differentiating the given equation with respect to yields:
Now solve for to obtain:
Set the equation equal to zero to find the maximum occurs at
Plug this back into the equation that we are trying to maximize and see that we are left with: .
The only answer choice that can be obtained from this equation is
Solution 5 (Lagrange Multipliers)
First, we move all the non-constant terms of the constraint to one side and assign it to the function : Since we are trying to maximize , we need to solve for and in the system We have that and To solve the original system, we can solve for and in terms of using our equations from the gradients, then substitute them into the first equation. We have that and . Substituting into the first equation, we have that Using the solutions of and in terms of that we found earlier, we have that Because we are trying to maximize this function, we will use the positive solution for . Therefore, after substituting, we have that the largest value of that satisfies is .
~qianqian07
Solution 6 (polar coordinates)
Completing the square gives the equation of a circle, Seeing that we would like to maximize we parameterize the circle using polar coordinates: Then, we have Since the desired answer is - Ultroid999OCPN
See also
1996 AHSME (Problems • Answer Key • Resources) | ||
Preceded by Problem 24 |
Followed by Problem 26 | |
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 • 16 • 17 • 18 • 19 • 20 • 21 • 22 • 23 • 24 • 25 • 26 • 27 • 28 • 29 • 30 | ||
All AHSME Problems and Solutions |
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.