Answer:
You need 4 weights, and the weights are 1; 3; 9 and 27.
Step-by-step explanation:
<em>Since the scale has two plates, we can place weights on either side and also the object so it can be balanced. </em>
This is a key part of the problem, it's not only on the other side of the scale, but on both sides.
Let's do the math now.
If i get two weights, 1 and 3. I can form this combinations.
Object of 1lb = 1
Object of 2lb + 1 weight = 3 weight.
Object of 3lb = 3 weight
Object of 4lb = 1 weight + 3 weight.
So what if i want to add the next weight and that weight to add me the maximum amount of objects. The weight would have to have a difference with the last object plus one. So if i grab 9. 9 minus 4 is 5. And that is a difference with the last object plus 1.
With a weight of 9, now i can add all the integers up to 13lb.
And the next step? Lets add one more. Keeping the last rule, the weight would have to have a difference with the last object plus one. So if i grab 27, 27 minus 13 is 14. And that is a difference witht the last object plus 1.
The sum of all the weights adds up to 40 pounds. And i can balance any integer in the middle.
The formula we are using is p – n = n + 1
Where p is the new weight. and n is the last object we weighted. And the sum of the weights goes up to the last object we can place on the scale, and in this case is 40.
. Because x2 is bigger than x1, you can think of f also becoming bigger, that is, f is increasing. The bigger the number x2, the bigger f becomes.
. The bigger the number x2, the smaller f becomes so f is "going down", that is, f is decreasing.