One number is 10 less than twice a second number.Find a pair of such numbers so that their product is as small as possible.
x-10
2x
2x(x-10)
2x^2-20x
-(-20/2(2))=5
2(5)^2-20(5)
50-100=-50
I think I am really lost
I agree you are lost.
one number=secondnumber-10
product=onenumber*(seconnumber )
product=(seconnumber-10)(secondnumber)
= Second^2-10Second
taking derivitives..
dProduct/dS=2S-10 set to zero
Second number=5
first number=15
I tried those answer previously and they are incorrect
by small, do you mean in absolute value? If so, then since
2*5-10 = 0,
5 and 0 will have a product of zero.
Otherwise, x(2x-10) has a minimum value of -12.5 when x = 2.5
Let's break down the problem step by step to find a pair of numbers that would result in the smallest possible product.
We are given two numbers: one is represented by "x" and the other is twice the first number, which can be represented as "2x". The problem states that one number is 10 less than twice the second number, so we can create an equation:
x = 2x - 10
Now, let's solve this equation to find the value of x:
x - 2x = -10
- x = -10
x = 10
So, one number is 10, and the other number is twice that, which is 20.
Now, let's find the product of these two numbers:
Product = x * (2x)
Product = 10 * (2*10)
Product = 10 * 20
Product = 200
Therefore, the product of these two numbers is 200.