Where in Interval does it Reach Absolute Maximum Value

cv2yanks13

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The volume V (in cubic centimeters) of 1 kg of water is very closely approximated by the function

v= 999.87 - (0.06426)T + (0.0085043) T^2 - (0.0000679)T^3

for temperatures ranging from 0 degrees Celsius to 30 degrees Celsius. At what temp on this interval does the volume reach its absolute maximum value?

I understand how to get the absolute max values from looking at the graph... but I need to show ALL algebraic and calculus work... I believe I need to find this by finding what the critical points are... I'm a little confused on how to find that (I'm uncertain if I have the correct answer)... can someone please assist me with this problem?? I truly need someone's help :)
 
The volume V (in cubic centimeters) of 1 kg of water is very closely approximated by the function

v= 999.87 - (0.06426)T + (0.0085043) T^2 - (0.0000679)T^3

for temperatures ranging from 0 degrees Celsius to 30 degrees Celsius. At what temp on this interval does the volume reach its absolute maximum value?

I understand how to get the absolute max values from looking at the graph... but I need to show ALL algebraic and calculus work... I believe I need to find this by finding what the critical points are.

Find the first derivative. Set it equal to zero and solve. This gives you relative maxima and minima.

Plug those values of T back into the original equation to find those relative maxima and minima values.

Check your boundary values at 0 and 30 also.
 
well i understand this... but I'm not sure what the derivative is supposed to look like...
is it supposed to be

0.0642+2(0.085043)t - 3(0.0000679)t ?????
 
well i understand this... but I'm not sure what the derivative is supposed to look like...
is it supposed to be

0.0642+2(0.085043)t - 3(0.0000679)t ?

The last t is squared:

dv/dt = 0.0642+2(0.085043)t - 3(0.0000679)t^2 = 0

Just solve the quadratic now using the quadratic formula. Make sure your solutions are inside the 0 to 30 domain.
 
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