if a and b are natural numbers and a>b then show that ( a²+b²), ( a²-b²) , (2ab) is a Pythagorean triplet. Find two Pythagorean triplets using any converilent value of a and b.
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Answer:
if a and b are natural numbers and a>b then show that ( a²+b²), ( a²-b²) , (2ab) is a Pythagorean triplet. Find two Pythagorean triplets using any converilent value of a and b.
answer to the question is :
1)
b² - a² = 2013
(b-a) (b+a) = 2013 = 3 * 11 * 61 prime factors.
b-a is smaller than b+a as and b are natural numbers.
Also b>a as RHS is positive.
Possibilities: b - a = 3 or 11 or 33
correspondingly b+a is 671 or 183 or 61
then b = 337 and a = 334 or b = 97 & a = 86 or b = 47 & a = 14
Product ab is least when a = 14 and b = 47 and is 658
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2)
S= x + y + z = 10 x, y, z ∈ N - { 0 } positive integers
A quick way is:
The given expression to be maximized : V = xyz + xy + yz + zx
S and V are symmetric in x, y and z. So maximum V is obtained when x = y = z = c.
Then c = 10/3 = non-integer. The possible values of x,y, and z are around 3.3.
Some possible combinations are: 3, 3, 4 ;;; 2, 4, 4 ;;; 5, 3, 2 ;;
x,y,z are : 3, 3, 4 => V = 69 this is the maximum.
x, y , z = 2, 4, 4 => V = 64
x, y, z = 5, 3, 2 => V = 61
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ANother way: 1<= x, y, z <= 8
V = 9 x y + 10 x + 10 y - x² y - x y² - x² - y²,
you can find dV/dx and dV/dy - partial derivatives and equate to 0. Then you find x = y = 3.3. at this value V is maximum.
So choose x, y, and z nearest to this value 3.3 and you get the maximum.
You can alternatives like above. x,y,z are 3,3,4 or 5,3,2 or 4,2,4 . For the combination 3,3,4 the value of the expression V is maximum and is 69.