Kohul asked in Science & MathematicsMathematics · 4 weeks ago

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?

Relevance
• 4 weeks ago

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.

• Puzzling
Lv 7
4 weeks agoReport

Non-negative includes zero. So x ≥ 0 and y ≥ 0. That's important for the maximum sum cases (0 and 30)

• Anonymous
4 weeks ago

a) 14 + 16 .....450

b) 29 + 1 .......842

• 4 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

• 4 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.