Question

Two positive numbers differ by 9 and the square of their sum is 144. Find the numbers.

Answer 1

This is a very simple problem in mathematics involving application of derivatives.

First let us assume the two numbers to be x and y .

From the question we see that we are given xy=144 .

Now we are asked to minimize the sum i.e., x+y .

Let us assume the minimum sum to be m .

So, by the question m=x+y -----1

Also xy=144 . So plugging the value of x or y (here we are plugging the value of y) in equation 1 we have,

m=x+144x

Now differentiating m with respect to x (w.r.t) we have,

dmdx=1−144x2 -----2

equating above to 0 we have,

1−144x2=0

=>x=+12,−12

Here we have the value of x.

Since we are asked to find the value of two positive numbers from the question we reject −12 .

Also we further verify that at x=+12 we have a minimum.

In order to do that we differentiate equation 2 one more time.

we get,

d2mdx2=288x3

we see by plugging the value of x=+12 in the above equation that the second derivative i.e., d2mdx2 is positive which further proves that at x=+12 we have a minimum.

Hence putting the value of x in xy=144 which was initially given to us we have the value of y=+12.

Hence we have the value of x as +12 and y as +12 and their sum is +24 which is minimum.

I hope you liked the explanation.

Answer 2

Answer:

let the numbers =x and y

x+9= y

(x+y)^2 = 144

(x+x- 9)^2 = 144

4x^2 - 81 =144

4x^2 = 144 +81

x^2 = 225/4

x = (225/4)^1/2

x= 15/2

x=7.5

then 7.5 +9 =y

y =16.5