Asked by Barb
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
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
Answers
Answered by
bobpursley
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
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
Answered by
Barb
I tried those answer previously and they are incorrect
Answered by
Steve
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
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
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