Find Max Sum Given Integers a,b,c,d

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The discussion focuses on maximizing the sum of four integers a, b, c, and d under the constraints a+b=c, b+c=d, and c+d=a, with the condition that b is greater than 0. By solving the linear equations, it is determined that a = -3b, c = -2b, and d = -b, leading to the sum a + b + c + d = -5b. The maximum value occurs when b is minimized at 1, resulting in a maximum sum of -5.

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a,b,c,d are integers

given :a+b=c, b+c=d , c+d=a

now if b>0, find max(a+b+c+d)
 
Last edited:
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Albert said:
a,b,c,d are integers

given :a+b=c, b+c=d , c+d=a

now if b>0, find max(a+b+c+d)

by solving we get

a = -3b,
c = -2b
d = -b

so sum = a + b + c + d = - 5b

it is largest when b ls smallest

b = 1 as b > 0 and integer

so largest sum = - 5
 
Last edited:
Albert said:
a,b,c,d are integers

given :a+b=c, b+c=d , c+d=a

now if b>0, find max(a+b+c+d)

The linear system of equations...

$$ a-c= - b$$

$$c- d= - b$$

$$a - c - d =0\ (1)$$

... has solution $a= - 3 b$, $c=- 2 b$, $d= - b$, so that is $a + b + c + d = - 5 b$ and the maximum is for b= 1, i.e. max (a + b + c + d) = -5...

Kind regards

$\chi$ $\sigma$

P.S. I didn't see that Kaliprasad solved before me... sorry!...
 

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