66The maximum value of 3x+4y is 73.
The idea is as follows. If we assume 3x+4y=k, then we are looking for largest k such that a point lying on the curve x2+y2-14x-6y-6=0 also satisfies 3x+4y=k.
The given curve is a circle with center (7,3) and radius 8. Obviously, k will be largest if the line 3x+4y=k is a tangent to the given circle. In that case the distance of the center (7,3) from the line 3x+4y=k should be 6. Thus
|3(7)+4(3)-k|√32+42 = 8
i.e. |33-k|=40
which gives k =-7 and k = 73. The maximum is obviously 73. This value is attained at the point of tangency which is (59/5, 47/5)
1how does the method to assume 3x+4y=k strike ?
Please tell me your thought process leading to this question this time :-)
1that is what they will test in IIT JEE.
coordinate geometry is itself a subject combining algegra and geometry.
of course, finding number of solutions or solutions to a set of equations can be solved by finding the number of intersecting points of figures formed by the corresponding equations.
for example if two 2nd degree equations with two variables are given, you can assume them to be circles and find intersecting points or solutions.
if two 2nd degree equations with three variables are given. you can assume them to be spheres and find intersecting points or circles which gives solution or a relation between the solutions.
NOTE : the algebric equations will be generally in terms of a,b,c.
you should carefully replace them by x,y,z(for convenience) and after solving points in terms of x,y,z you can take them as solutions for a,b,c respectively
11Interesting.....I thought this was the std way of tackling such problems....champ!
This should not be 2 unfamiliar.....
341A slight variant on what kaymant sir has done:
Notice that \frac{|3x+4y|}{5} measures the distance of the point (x,y) from the line 3x+4y = 0.
So, the problem asks us what is the greatest distance of any point on the circle
(x-7)2 + (y-3)2 = 82 from the line 3x+4y = 0.
If you draw a diagram it is easy to see that this is nothing but radius + distance of centre from 3x+4y = 0, which is 8 + 33/5 = 73/5
Hence the maximum value is 5 X 73/5 = 73
341
Hari Shankar
·2009-08-23 22:16:18
Another method would be to parameterize every point as x = 7 + 8 cos θ and x = 3 + 8 sin θ and minimize 3x+4y
If you know inequalities you can settle this most easily as from Cauchy-Schwarz inequality we have
(3^2 + 4^2) [(x-7)^2 + (y-3)^2] = 25 \times 64 \ge [3(x-7) + 4(y-3)]^2 = (3x+4y-33)^2
\Rightarrow 40 \ge 3x+4y-33 \ge -40 \Rightarrow 73 \ge 3x+4y \ge 7