the sum of 2 nonnegative numbers is 30. Find numbers a) if the sum of their squares is a minimum, b) if sum of their squares is a maximum ?
Using calculus,how would you do this?
- llafferLv 74 weeks agoFavorite Answer
The sum of two non-negative numbers is 30:
x + y = 30 where x > 0 and y > 0
Find the numbers if the sum of their square is a minimum:
We can set up this equation:
s = x² + y²
If we solve the first equation for x in terms of y we get:
x = 30 - y
Substitute and simplify into the second equation to get:
s = (30 - y)² + y²
s = 900 - 60y + y² + y²
s = 2y² - 60y + 900
The value of "y" that gives "s" a minimum can be found by solving for the zero of the first derivative:
ds/dy = 4y - 60
0 = 4y - 60
-4y = -60
y = 15
Now that we have y, solve for x:
x = 30 - y
x = 30 - 15
x = 15
Both numbers are 15 to make "s" the minimum value of 450.
- Anonymous4 weeks ago
a) 14 + 16 .....450
b) 29 + 1 .......842
- Michael ELv 74 weeks ago
Let x be one of the numbers. (30-x) is the other number.
The sum of their squares is x^2 +(30-x)^2
= x^2 + 900 - 60x + x^2
= 2(x^2) - 60x + 900
At both the maximum and the minimum, the slope will be 0 or the point will be at a boundry.
Slope at x is 4x - 60
- PuzzlingLv 74 weeks ago
Let x be the first non-negative number (x ≥ 0)
Let 30 - x be the second non-negative number (30 - x ≥ 0)
If you check both the inequalities, you see that both numbers are between 0 and 30 inclusive.
Intuitively, you should be able to figure out the minimum is when both numbers are equal. And the maximum would be when one is 0 and one is 30.
But let's be more formal and solve it using calculus.
Write a formula for the sum of their squares.
f(x) = x² + (30 - x)²
f(x) = x² + 900 - 60x + x²
f(x) = 2x² - 60x + 900
If you graph this (or remember algebra II), you'll find it is an upward facing parabola and you could figure out the vertex is a minimum when x = 15. From this you could get the answer.
Again, let's forget the easier route and continue using calculus.
Take the derivative:
f'(x) = 4x - 60
Set it to zero:
f'(x) = 4x - 60 = 0
4x = 60
x = 15
But you don't know if this is a maximum or minimum. You could take the second derivative.
f"(x) = 4
Since this is positive that means the curve is concave up and so x = 15 would be a minimum.
You'd then have to look at the graph from 0 to 30 to see that the maximums are at the endpoints.
15 and 15 --> minimum
0 and 30 (or 30 and 0) --> maximum