Find two numbers whose difference is 30 and whose product is a minimum.

Answer 1

Write y = x + 30, or y − x = 30.

Next, if y = x + 30, then xy or x(x + 30) or x² +30x

gives the product of the two numbers x & y.

Now write f(x) = x² +30x. By The First Derivative Test, f'(x) = d(x² +30x)/dx or 2x + 30 gives a critical point at x = -15 (to obtain 2x + 30 equal to 0).

Then The Second Derivative Test gives f''(x) as d²y/dx² equal to d(2x + 30)/dx or 2 which is greater than zero. This positive value of 2 for f''(-15) indicates that f(x) = x² +30x has a lowest point or minimum for

(x,[x² +30x]) at (-15,[(-15)² +30(-15)]) or (-15,-225).

Since -225 is obtained by (-15 times 15) and (15 minus -15) gives a difference of 30, the two numbers sought would be (x,y) equal to (-15,15).

Answer 2

Calculus is not always the most direct.

z = x² - 30

From Algebra minimum is at x = -(-30) / 2(1)

Answer 3

We can think of a function z(x,y) = xy, where x and y can be the two numbers. We can add the constraint that x - y = 30 (here we're saying x is the larger number arbitrarily).

Like any function, when we want to find its minimum we take its derivative, set it equal to zero and solve for the variable values at which this minimum occurs. This is annoying if z is a two variable function, but we can easily make it single variable by recognizing that y = x - 30. Therefore:

z = x(x-30)

z = x² - 30x

dz/dx = 2x - 30

2x - 30 = 0

x = 15

This means y = -15, so the two numbers are -15 and 15.