Find the maximum value of 3x + 4y if x^2 + y^2 = 6x - 4y - 4
@Calvin4ever said:Find the maximum value of 3x + 4y if x^2 + y^2 = 6x - 4y - 4
the answer for max(3x+4y) is 13..and its at the point (3,1)..x^2+y^2-6x+4y+4=0 is a circle having centre(3,-2) and radius 3.
for any given circle any line can be as fsar as possible if it's a tanget
so for 3x+4y-d=0 distance from center is
3*3+4*-2-d/5=3
d=16 as ans
for any given circle any line can be as fsar as possible if it's a tanget
so for 3x+4y-d=0 distance from center is
3*3+4*-2-d/5=3
d=16 as ans
@Calvin4ever said:Find the maximum value of 3x + 4y if x^2 + y^2 = 6x - 4y - 4
approx 16 ??
1 + 21/root(2)
@ishu1991 said:the answer for max(3x+4y) is 13..and its at the point (3,1)..x^2+y^2-6x+4y+4=0 is a circle having centre(3,-2) and radius 3.for any given circle any line can be as fsar as possible if it's a tangetso for 3x+4y-d=0 distance from center is3*3+4*-2-d/5=3d=16 as ans
Can you elaborate a bit to the darkened line
What is the area of the largest equilateral triangle that can be cut from a circular piece of paper or radius 1 cm?
Is there any good approach to find exact rate % for compound interest other than checking for options ??? i mean if r % annually change is 100 % for 2 years sort of ques...it is really troubling when v do it for 3 successive changes .......
@catahead said:What is the area of the largest equilateral triangle that can be cut from a circular piece of paper or radius 1 cm?
3_/3 /4 ?
Side of triangle should be _/3
Area=_/3/4 * (_/3)^2=3_/3 /4
@catahead said:What is the area of the largest equilateral triangle that can be cut from a circular piece of paper or radius 1 cm?
3rt(3/4) ?
a/rt3 = r
a/rt3 = 1
a = rt3
area = rt3/4*a^2 = rt(3/4)*(rt3)^2 = 3rt(3/4)
@Swavi91 said:@saurav205 wat was ur approach??..kindly explain.
abc/4R = Area.., R =1. It should be equilateral., hence, a=b=c.
Area will be (_/3/4)a^2..
Hence, area will be 3*/3/4.!
@Swavi91 said:@saurav205 wat was ur approach??..kindly explain.
abc/4R = Area.., R =1. It should be equilateral., hence, a=b=c.
Area will be (_/3/4)a^2..
Hence, area will be 3*/3/4.!
@catahead
Shld be (3*_/3)/4..
Height of the triangle(AD) = (_/3)/2*a [a=Side of the Equilateral Triangle ABC]
Now, Let O be the center of the circle..Thus, OC=r cm..OB= AD-r..BC=a/2..
Thus, r^2 = a^2/4 + [(_/3)/2*a - r]^2
=>a= _/3r cm..
Thus, a= _/3 cm..n the area of the Triangle= (_/3)*3/4 cm^2..
in a class the ration of number of boys to girls is 7:3.Each boy is given only a 50 paise coin and each girl is given a 75 paise coin.The difference in the amount present with boys and girls is 3.75.how many coins should boys and girls exchange so that amount with boys become twice the amount with the girls.
@Calvin4ever said:Find the maximum value of 3x + 4y if x^2 + y^2 = 6x - 4y - 4
there are 2 ways of doing it...
1) by cauchy inequality..
we have (x-3)^2 + (y+2)^2 = 9
{(x-3)^2 + (y+2)^2}(3^2+4^2) >= (3(x-3) + 4(y+2))^2
25(x^2+y^2-(6x-4y-4)+9) >= (3x+4y-17)^2
25(x^2+y^2-(x^2+y^2)+9) >= (3x+4y-1)^2
25*9 >= (3x+4y-1)^2
3x+4y-1 =3x+4y =
2) by applying the logic that the farthest point from the circle is that point where the tangent touches it... (by this method we get both, the max and the min value)
(x-3)^2 + (y+2)^2 = 3^2 is our equation of the circle with center at (3,-2) and radius = 3
using distance from center = radius
|{3*3 + 4*(-2) - k}|/rt(3^2+4^2) = 3
+/- (1-k) = 15
14 =hence max is 16....
@pakkapagal said:in a class the ration of number of boys to girls is 7:3.Each boy is given only a 50 paise coin and each girl is given a 75 paise coin.The difference in the amount present with boys and girls is 3.75.how many coins should boys and girls exchange so that amount with boys become twice the amount with the girls.
ans 4?
If the eight-digit number N = 76542a3b is divisible by 18, find the number of possible values of N.
@shinoda said:If the eight-digit number N = 76542a3b is divisible by 18, find the number of possible values of N.
ans 5?