SOLUTION: DURING A MANUFACTURING PROCESS A METAL PART IN A MACHINE IS EXPOSED TO VARYING TEMPERATURE CONDITION.THE MANUFACTURER OF THE MACHINE RECOMMENDS THAT THE TEMPERATURE OF THE MACHINE

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Question 1120311: DURING A MANUFACTURING PROCESS A METAL PART IN A MACHINE IS EXPOSED TO VARYING TEMPERATURE CONDITION.THE MANUFACTURER OF THE MACHINE RECOMMENDS THAT THE TEMPERATURE OF THE MACHINE PART REMAIN BELOW 135F. THE TEMPERATURE,T IN DEGREE FAHRENHEIT X MINUTES AFTER THE MACHINE IS PUT INTO OPERATION IS MODELED BY THE EQUATION. T=-0.005X SQUARE + 0.45x + 125.
a. Tell whether the temperature of the part will ever reach or exceed 135 Fahrenheit. Using discriminant quadratic equation
b. If the machine is in operation for 90 minutes before being turned off, how many will the temperature of the part be 134 fahrenheit
Answer by Theo(13342) (Show Source):
You can put this solution on YOUR website!
the equation is T = -0.005 * x^2 + .45 * x + 125.

the equation is in standard form of ax^2 + bx + c

a = -.005
b = .45
c = 125

the maximum value of T is when x = -b/2a.

that would be when x = -.45 / (2*-.005)

simplify to get x = -.45 / -.01

solve for x to get x = 45

when x = 45, T = -.005 * x^2 + .45 * x + 125 = 135.125 degrees.

yes, the part will reach or exceed 135 degrees.

if the part is in operation for 90 minutes, the part will have reached 134 degrees two times.

you can replace T with y and graph the equation of y = -.005 * x^2 + .45 * x + 125.

the graph looks like this:

from the graph, you can see that the part will reach or exceeed 134 degrees when x = 30 through x = 60.

that's a period of 30 minutes.

from the graph, you can also see that the maximum value of the temperature is 135 degrees.

the part will be 125 degrees when x = 0 and when x = 90, according to the formula.

the points at which the temperature reaches 134 degrees was found by taking the equation of y = -.005 * x^2 + .45 * x + 125 and replacing y with 134.

you get 134 = -.005 * x^2 + .45 * x + 125.

subtract 134 from both sides of this equation to get -.005 * x^2 + .45 * x - 9 = 0

factor this equation to get x = 30 or x = 60.

i used an online quadratic equation solver at https://www.mathsisfun.com/quadratic-equation-solver.html

the results from that quadratic equation solver are shown below:

if the quadratic equation solver is unavailable, you would solve using the quadratic formula.

that formula is referenced below:

http://www.purplemath.com/modules/quadform.htm