Respuesta :

Let P be the product, x be the first number and y be the second number
therefore:
x-y=68........i
P=xy....ii
from i
x=68+y
hence
P=y(68+y)=y^2+8y
P=y^2+8y
The minimum product will occur at point where:
dP/dy=0
thus
dP/dy=2y+8=0
thus
y=-4
thus the value of x will be:
x=68-4=64
thus the product will be:
-4*64=-256 sq. units

We want to find two whole numbers whose difference is 68 and whose product is minimum.

The two numbers are 34 and -34, and the product is:

P = -1,156

Let's say that our two numbers are A and B, such that A is the smaller one.

We know that the difference must be 68, then we have:

B - A  = 68.

And the product between these two numbers can be written as:

P = B*A

If we isolate one of the variables in the first equation, we get:

B = 68 + A

Now we can replace that in the product to get;

P = (68 + A)*A = 68*A + A^2

We want to minimize this, notice that this is a quadratic polynomial of positive leading coefficient, meaning that the minimum is at the vertex.

And for the general quadratic polynomial:

y = a*x^2 + b*x + c

The vertex is at:

x = -b/2a

Then in this case the vertex is at:

A = -68/(2*1) = -34

The value of B is given by:

B = 68 + (-34) = 34

The product is:

P = 34*(-34) = -1,156

If you want to learn more, you can read:

https://brainly.com/question/3496582