Question 1209912: Let x and y be positive real numbers. If x + y = 1, then find the maximum value of xy + y^3.
Found 3 solutions by greenestamps, math_tutor2020, mccravyedwin:
Answer by greenestamps(13200)   (Show Source):
Answer by math_tutor2020(3817)  (Show Source):
You can put this solution on YOUR website!
x+y = 1 rewrites to y = 1-x
xy + y^3 = x(1-x) + (1-x)^3
You can optionally expand everything out, but it's not really necessary.
Graphing that with a calculator such as GeoGebra or Desmos would lead to this
Let's focus on the portion where x > 0 since x is a positive real number.
There is an open hole at x = 0 since 0 isn't a positive number.
Think of this open hole as a pothole you cannot drive over on a road.
Despite not being able to drive over this hole, we can get arbitrarily closer and closer to it.
We can try x = 0.1, then x = 0.01, then x = 0.001 etc
As x gets closer to 0 from the right, we steadily approach f(x) = 1 from below.
We never actually reach f(x) = 1 itself.
Here is a table of select values
| x | f(x) = x(1-x)+(1-x)^3 | | 0.1 | 0.819 | | 0.01 | 0.980199 | | 0.001 | 0.998002 | | 0.0001 | 0.9998 | | 0.00001 | 0.99998 |
The decimal values are approximate.
As you can see, there is no maximum value to xy + y^3 when x+y = 1 and x,y are positive real numbers.
Answer by mccravyedwin(407)  (Show Source):
You can put this solution on YOUR website!
Since x and y are positive real numbers, and x + y = 1, then x can only go up
from 0 to 1, while y goes down from 1 to 0. Look at this table of values, As we
see, xy + y3 goes down from 1 to 0.
So its maximum value ought to be 1 when x is 0 and y=1. But 0 isn't positive.
So I guess we have to say is that xy + y3 is bounded. Its least upper
bound is 1. But it has no maximum. Who makes up these silly problems?
x y xy + y3
0.0 1.0 1.000 <-- forbidden maximum value
0.1 0.9 0.819
0.2 0.8 0.672
0.3 0.7 0.553
0.4 0.6 0.456
0.5 0.5 0.375
0.6 0.4 0.304
0.7 0.3 0.237
0.8 0.2 0.168
0.9 0.1 0.091
1.0 0.0 0.000 <--- forbidden minimum value.
Edwin
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