How do you solve this problem using calculus?

Question

I have solved this problem by using the standard algebraic approach, how do you solve this using calculus methods?

1) If the sum of the cubes of the roots of x^2-bx+10=0 is 6b, compute the greatest possible value of b.

Answer 1

if U want to tally the answer, write your answer. I am doing it for the 2nd time. I deleted it earlier. Bcause while asking you people show as if sky will fall and then after trying a new question answerer has to depend on you for talloed or not tallied with your answer
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x^2-bx+10=0 is

let m n be 2 roots
m+n = b and mn =10-----------(A)
1 + dn/dm = db/dm---------(B)

given 6b = m^3 + n^3 ----------(C)
differentiate for greatest

6 db/dm = 3 m^2 + 3n^2 dn/dm ....(use (B)
6 db/dm = 3 m^2 + 3n^2 [db/dm - 1]
db/dm = 3 m^2 - 3n^2 / [6 - 3 n^2]
for greatest db/dm =0 >>> 3 m^2 - 3n^2 =0
m = n or m = -n

m=n***************
6(b)max = m^3+m^3 = 2 m^3
also m n =10 >> m^2 =10 or m =+- sqrt (10)

so (b)max = (m^3/3) = +- (10)^3/2 / 3

(b)max = +- 10.54 ****************
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m = - n is not allowed because
(b)max = m^3 - m^3 = 0

Now what is your answer

Answer 2

This isn't a calculus problem, as there are only 3 possible values of b. (i.e., there is no function representing b on which to take a derivative and set = 0. b is not a function of anything. It is uniquely defined by the problem.)
In a quadratic ax^2 + bx + c=0, the sum of the roots is -b/a and their product is c/a

Let the roots of your equation be m and n.
m+n = b
mn = 10

(m+n)^3 = b^3
m^3 + 3mn(m+n) + n^3 = b^3
m^3 + 3*10b + n^3 = b^3
m^3 + n^3 = b^3 - 30b
6b = b^3 - 30b
b^3 - 36b = 0
b(b^2-36) = 0
b = 0 or b = ±6

so b = 6 is the solution.

{If the problem had stated that the sum of the cubes of the roots was S, and you wanted to find the value of b that would maximize S, or find the maximum value of S, then that would be a calculus problem.}

Answer 3

I'm in Calculus II at the moment and don't think that there is a calculus approach to this problem. There is no calculus method for finding the roots of a function, cubing a root, or setting the sum of those equal to 6b

I think that calculus would only complicate the problem.

Answer 4

If x and y are the roots
x^3+y^3 =(x+y)(x^2-xy+y^2) = (x+y) [(x+y)^2-3xy]
Now we can express all the terms remembering that
x+y= b and xy = 10
so b( b^2 -30) = 6b
b(b^2-36)=0 b=0 and b=+-6 so the greatest possible value of b is
b=6
There is no function to maximize so no use of calculus methods

Answer 5

Car A leaves SF at 50 mph Car B leaves DEN 1 hr later at 60 mph Don't you need to know the distance between SF and DEN? Visualize the problem this way- in 2 hrs A has traveled 100 mi and B has traveled 60 miles When their combined distance traveled equals the distance between SF and DEN they meet. Distance A traveled = t x 50 Distance B traveled = (t-1) x 60 t x 50 + (t-1) x 60 = distance between SF and DEN 50t+60t - 60 = distance between SF and DEN Solve for t (time A traveled; B traveled 1 hr less) Compute the distance from DEN of A (at t) and B (at t-1). The trick part of the question is how you define "meet". If it is front bumper to front bumper, then the car from DEN will be closest to DEN. If "meet" means they are side by side, then it still is a trick question because they would both be the same distance from DEN.

Answer 6

To find the max/min values of an equation, you take the derivative.

Answer 7

From what I remember out of calc, you have to use derivates with respect to b.

Answer 8

Come on you know it.i saw u do it at home.your in 11TH grade.COME ON!
people,don't awnser this ?.hes trying to ask you if you Know it 2.