Question

using differentiation, how do you solve this?

find two positive numbers whose product is 64, and whose sum is a minimum. (the answer should be: (8,8) )

Answer 1

Let the two positive numbers be x and x/64 Now form y=x+x/64
Take derivative of y with respect to x Now carry on...........

Answer 2

why is the other number x/64?

Answer 3

because the product of x and x/64 is 64. Check it out by yourself

Answer 4

Wouldn't it be x and 64/x

Answer 5

Sorry yes x and 64/x

Answer 6

What a shame!

Answer 7

then, why not 64/x?

Answer 8

Now y=x+64/x
And differentiate

Answer 9

Sorry! My blunder, What the hell!

Answer 10

it's alright.

Answer 11

Now can you work out from here

Answer 12

so, it will be 1 + (-64/x^2) =y when differentiated?

Answer 13

Yes

Answer 14

Yeap, now equate it with 0

Answer 15

You will get x=8 and -8

Answer 16

x=8 and x =-8

are they the answers?

Answer 17

but the number should be positive so choose 8

Answer 18

Hence the numbers are 8 and 64/8=8

Answer 19

because a minimum is > 0? is that it?

Answer 20

A maximum or minimum is when the first derivative is equal to zero.

Answer 21

Second derivative will be greater than zero check it out by yourself

Answer 22

Also the first derivative is always zero for minimum and maximum

Answer 23

\[xy=64\] <-two positive numbers whose product is 64
Let S be the sum of x and y
\[\ S=x+y\]
so we know xy=64 => y=64/x

so we can write S using one variable

\[\ S=x+\frac{64}{x}\]

now S' needs to be found since we are trying to minimize

\[\ S'=1-\frac{64}{x^2}\]

now we need to set S'=0

\[\frac{x^2-64}{x^2}=0\]

8,-8, and 0 are critical numbers but we throw out -8 and 0 since x>0 (remember we looking for numbers and also if x=0 then the product would not be 64 it would be 0)

so we have x=8 => y=64/8=8

Answer 24

positive numbers*