Question

the product of two whole numbers is 36. What is the greatest possible sum the number can have?

Answer 1

The divisors of 36 are 1, 2, 3, 4, 6, 9, 12, 18, and 36. We can see that when paired properly the pair with the greatest sum is 36+1=37

Answer 2

@dacelin_naomie

Answer 3

xy=36
S=x+y
We want to maximize S, correct?

Answer 4

We are given that \(a\times b=36\) and need to maximize \(a+b\).
Applying The Method of La Grange Multipliers:
$$
g(a,b,\lambda)=a+b+\lambda(a\times b-36)\\
g_a=1+\lambda a=0\implies a=\cfrac{-1}{\lambda}\\
g_b=1+\lambda b=0\implies b=\cfrac{-1}{\lambda}\\
g_\lambda=a\times b-36\implies \cfrac{1}{\lambda^2}=36\\
\lambda=\pm\cfrac{1}{6}\\
a=\pm6\\
b=\pm6\\
$$
So, the greatest possible sum the number can have is \(6+6=12\).

Answer 5

hey @myininaya am i correct?

Answer 6

@ybarrap 36+1=37 ----------------------36*1=36

Answer 7

Hmmm...I think you minimized instead of maximized @ybarrap

Answer 8

And yes @Uditkulka looks good.

Answer 9

I agree, looks like I minimized. @Uditkulka 's solution looks right.

Answer 10

I wanted to do some calculus too @ybarrap :(