Question

Find two numbers whose difference is 10 and whose product is a minimum.​

Answer 1

Answer:

Step-by-step explanation:

let the numbers be x and y,let y>x

y-x=10

y=x+10

product P=xy=x(x+10)=x²+10x

$\frac{dP}{dx} =2x+10\\\frac{dP}{dx} =0,gives\\2x+10=0\\2x=-10\\x=-5\\\frac{d^2P}{dx^2} =2>0 ~at~x=-5$

∴P is minimum at x=-5

y=x+10=-5+10=5

numbers are -5,5

Answer 2

Answer:

Answer: {-5, 5}. Product is -25, which is the minimum.

Step-by-step explanation:

let a, b denote the two numbers. We know that b-a=10.

We are looking for a minimum over the product a*b.

One can minimize this using derivatives. In case you have not yet had derivatives, you can also use the vertex of a parabola (since the above is a quadratic form):

The minimum is at the vertex a=-5 and so b=5

Their distance is 10, and their product attains the minimum value of all possiblities -25.