In the event of a power failure, a computer model estimates the temperature T (in degrees

Celsius) of a food processing plant’s freezer to be Tt= ((2t^2)/(t+2) ) -4, where t is the time in hours

after the power failure. If the power failure happened at 4pm, find the time when the ice cubes

in the freezer started to melt.

reshown

Ice cubes melt at temperature above 0 degree celsius.

Or  ((2t^2)/(t+2) ) -4 =0

Or 2t^2 = 4t + 8 =0

Or 2t^2 - 4t -8 = 0

or t^2 -2t - 4 =0

On solving the following quadratic equation using Quadratic formula, t = -b +- sqrt(b^2 -4ac)/2a, we get:

t =  1 - sqrt(5)  or   t = 1 + sqrt(5)

Neglecting t = 1- sqrt(5) as time doesnot flow backward.

Therefore, the ice cubes will start melting at  4 + 1 + sqrt(5) = 5 + sqrt(5)

or it will start melting at 08: 14 : 10 pm

by Level 7 User (26.4k points)

Ice cubes melt at 0°C so:

0=2t²/(t+2)-4,

t²/(t+2)=2,

t²=2t+4,

t²-2t=4,

t²-2t+1=5,

(t-1)²=5,

t-1=±√5,

t=1+√5 (reject the negative result),

t=3.236hr approx after 4pm.

The ice cubes will melt around 7:14pm.

by Top Rated User (816k points)