How would I go about Finding two numbers whose difference is 100 and whose product is a minimum?

Answer 1

Let x be the larger number
Let x - 100 be the smaller number
The difference between the two is 100.

The product would be x(x-100) = x² - 100x.

If you take the derivative, you get 2x - 100.

Set this to zero:
2x - 100 = 0
2x = 100
x = 50

Alternatively, if you aren't familiar with calculus, turn the equation into vertex form for a parabola. Do this by completing the square:
x² - 100x = 0

Take the coefficient on the x-term (-100), divide it by 2 (-50), and square it (2500). Add this to both sides:
x² - 100x + 2500 = 2500
(x - 50)² = 2500
(x - 50)² - 2500 = 0

This is vertex form a(x - h)² - k = 0 and the vertex will be at the point x = h or x = 50.

So either way you figure it, the numbers are 50 and -50. The product is -2500 which is minimum.

Answer 2

If x - Q = 100, then Q = 100 + x

Substitute this into the following equation:

F(x) = x*Q <----this is the eqation for which you want to find a minimum

F(x) = x *(100 + x)

F(x) = x^2 + 100x

Graph that equation, and look for its lowest point. The lowest point is at x = -50, f(x) = -2500

Since Q = 100 + x, then Q = 100 + (-50) = 50

50 - (-50) = 100, and the product--being -2500--is a minimum.

Answer 3

think of the function in 3-d space, z = x * y

it is worth 0 for all x=0 or y=0. Take examples. For x=1, z=y so it is just a line with slope 1, right. For x=2, same thing but slope 2. So you've got ever steeper lines as x's go up. And the slopes are just inverted for negative x's.

And it makes the classical shape of the x^2 parabola for x=y

partial derivative vs. x is equal to y, so will be equal to 0 only at y=0 where you have a saddle point

ditto for partial derivative w respect to y


but now you've got a contraint: y=x+100, which prevents you from hitting that saddle point. So you could go for ever more negative numbers, and you'd get ever more negative z's

now let's assume you forgot to mention that you were limited to POSITIVE x's and y's. In which case you'll get the minimum, under the constraint y=x+100, for the smallest allowed x, i.e. 0.


hope this helps

Answer 4

Let x and y be the numbers, with x being the larger. Then x-y = 100, and you want to minimize xy. Replace y by x-100, and minimize x(x-100).

To do this, differentiate it, and get 2x-100. Set it equal to 0, and you get x=50. By computing the second derivative, which is 2, you can confirm that this critical number corresponds to a local minimum.

Also, this is the absolute minimum, because as x -> inf or -inf, the product approaches infinity, so x = 50 and y = -50.

Answer 5

x - y = 100, so y = x - 100 the product is x*y = x*(x-100) = x^2 -100x since you have an expression, take the derivative and let it equal zero: 2x - 100 = 0 so x = 50 and y = -50

Answer 6

x - y = 100
I know that one of the numbers must be negative, because the product will be negative that way. Now, what too numbers will give the biggest product? (large negative numbers will be the minimum)

Lets try some examples
(-1) * 101 = -101
(-50) * 150 = -7500
(-51) * 151 = -7 701
(-3 000) * 3 100 = -9 300 000
(-6 000 000) * 6 000 100 = -3.60006 × 10^13
i see that i can pick arbitrary large numbers and come out with a smaller answer. There is no minumum answer.

Answer 7

let the two numbers be x+100 and x .

d( x^2+100)/dx=0

2x+100=0
and x=-50 and another one is 50 over

Answer 8

x - y = 100
x = y + 100

y(y + 100) = y^2 + 100y

y = (-100)/(2(1))
y = -50

x - y = 100
x - (-50) = 100
x + 50 = 100
x = 50

ANS : (50,-50)