Find two positive numbers whose product is 196 and whose sum is a minimum.?

OK
So you want a *b = 196 where a+b is minimum.
196 = 196 *1 and the sum would be 197
196 = 98 *2 and the sum would be 100
Notice that as the lower factor increases, the sum decreases. Notice that, as well, there will reach a point of minimum, and then the second sum would increase as the second number became larger (think of the two numbers in reverse).
The absolute minimum would occur at the sqrt of 196 if it is a perfect square – and it is – 14.
So the answer is 14,14 and the minimum sum is 28.
Notice that the factors of 196 are 7*7*2*2 and the factor table looks like this:
196, 1
98, 2
49, 4
28, 7
14, 14 and then you can reverse the numbers.
Notice that at 14, 14 you will have the lowest sum. The second lowest combination would be 28,7 for s sum of 35.
Hope that helps.

Also Read :   Enter the correct value so that each expression is a perfect-square trinomial. x2 – 10x + 25 x2 + x + 36

a*b=196 s=a+b=a+(196/a) ds/da = a million + 196*(-a million/(a^2)) For sum to be minimum (ok, must be optimal) ds/da = 0 For ds/da = 0, a million = 196/(a^2) a^2 =196 a=14<<< b=196/a = 14<<< (as the different answerer stated- yet i have given an evidence.)

Factoring is OK but you can also use a little calculus:
n*m = 196 … n =196/m
s = n + m = 196/m + m
ds/dt = 0 = -196/m^2 + 1
m^2 = 196
m = 14

2 * 2 * 7 * 7 = 196
14 * 14 = 196
14 + 14 = 28
The other combinations give a higher sum :
2 * 98 => sum = 100
4 * 49 => sum = 53
7 * 28 => sum = 35
(14 ; 14) is the solution

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