By what factor should I increase all dimensions of my slingshot to increase its maximum distance by 50%?

Question

(the projectile is increased in the same proportion)
(ignore air resistance)

"Thus x is multiplied by 2 and k by 4"

Not really.
Cross-section of rubber multipled by 4, but because the length is doubled, k is multiplied only by 2

Answer 1

A length factor of 2 multiplies cross-section by 4, volume and mass by 8. Since length is doubled, the slingshot can be pulled twice as far back. Thus x is multiplied by 2 and k by 4, so energy (1/2 kx^2) is multiplied by 16. With projectile mass increased by 8, specific energy (1/2 v^2) increases by 2, and range increases by sqrt(2). This is a 41% increase, pretty close to 50%. For a 50% increase you want to scale the slingshot by 1.5^2 = 2.25.
EDIT: Thanks for the correction. Of course 8 * as much rubber = 8 x the energy storage. So you can't increase the range!

Answer 2

When a projectile is fired at 45º for maximum range, the distance is proportional to v². You can verify this by searching under "maximum range" projectile under Google.

I'll make the bad, but not unreasonable assumption, that the kinetic energy of the projectile, KE = ½mv², is equal to the work done in pulling back the slingshot, W = ½kx².

If ½mv² = ½kx², then v² = kx²/m. If you've increased the projectile mass "in proportion" by 50%, you halve v². To double the original range with the bigger projectile, you'll need to increase kx² by a factor of four. This can be done by doubling how far you pull the slingshot back or by doubling the width and thickness of the elastic to increase k by a factor of four.

All things considered, I'd double my slingshot size.