SOLUTION: The minimum cost of maintaining an overhead crane depends on the number of hours the crane is in operation. The cost is given by the relation C=6t^2

SOLUTION: The minimum cost of maintaining an overhead crane depends on the number of hours the crane is in operation. The cost is given by the relation C=6t^2 - 36t +154, where C represents

Question 1209023: The minimum cost of maintaining an overhead crane depends on the number of hours the crane is in operation. The cost is given by the relation C=6t^2 - 36t +154, where C represents the cost in hundreds of dollars and t represents the time in hours that the crane has been operated. a) What is the minimum cost of maintaining the crane? (b) Find the number of hours the crane operates for this minimum maintenance cost.
Answer by math_tutor2020(3816) (Show Source): You can put this solution on YOUR website!

I'll use x in place of t and use y in place of C.
The equation is y = 6x^2 - 36x + 154.
Compare this to the standard quadratic template y = ax^2+bx+c
a = 6, b = -36, c = 154

a = 6 is positive, so we have a parabola opening upward to indicate the vertex is the lowest point.
The vertex is located at (h,k)
h = -b/(2a)
h = -(-36)/(2*6)
h = 3
This is the x coordinate of the vertex.
It's the number of hours needed to reach the lowest cost.

Plug this x value back into the original equation to find its paired y value.
y = 6x^2 - 36x + 154
y = 6*3^2 - 36*3 + 154
y = 100
Therefore the vertex is located at (3,100).

Note that when completing the square, y = 6x^2 - 36x + 154 turns into the vertex form y = 6(x-3)^2 + 100.

y represents the cost in hundreds of dollars.
Whatever the y value is, multiply by 100.
Something like y = 2 represents 2*100 = 200 dollars
y = 100 represents 100*100 = 10,000 dollars.

Answers:
a. $10,000
b. 3 hours

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