Question

Find two numbers differing by 46 whose product is as small as possible.

Answer 1

Let the numbers be x and y.  They differ by 46. Then x = y + 46.

Their product is P = xy.  Since x = y + 46, P = xy = (y + 46)y, or

P = y^2 + 46y.  You could graph this and then identify the coordinates of the vertex, which would give you the minimum value of P.

Or you could differentiate P(y) with respect to y, set the result = to 0, and solve for y:  

2y + 46 = 0; y = -23.  x = y + 46, or +23.

The vertex of the graph of this parabola represents the minimum value of the product P.  It is (23, 46), and 46 is the smallest possible product here.

Answer 2

Answer:

The numbers are 23 and -23.

Step-by-step explanation:

If the two numbers are x and y, we can say that the product is

$P=x\cdot y$

We need to eliminate a variable. To do that we use the fact that the difference between the two numbers has to be 46.

$x-y=46$

So we can say $y=x-46$ and plug that into the equation for P

$P=x\cdot (x -46)=x^2-46x$

From the information given we want the product P to be minimized. For this we find the derivative

$\frac{d}{dx}(x^2-46x) =\frac{d}{dx}\left(x^2\right)-\frac{d}{dx}\left(46x\right)=2x-46$

Next, we find the critical points, we set the above equation equal to zero and solve for x

$2x-46=0\\2x-46+46=0+46\\2x=46\\\frac{2x}{2}=\frac{46}{2}\\x=23$

Since x = 23 is the only critical number, we can conclude that there’s actually an absolute minimum at x = 23

Thus, $x = 23$ and $y = 23-46=-23$