calculus problem involving water density and maximum values -- HELP!!?

Question

Between 0degrees celcius and 30degrees celcius, the volume V (in cubic centimeters) of 1 kg of water ata a temperature T is given approximately by the formula

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

Find the temperature at which water has its maximum density.


I know the answer is around 3 degrees celcius just because i remember it from physics but i dont know how to get the answer using calculus (max and min values)

Answer 1

to obtain the answer:

differentiate the equation with respect to x to find V'

make V' equal to 0

solve that equationto find x

this will give you 2 answers one is right one is wrong they can be tested by the second derivative test or the sign rule

Answer 2

V = 999.87 - 0.06426 T + 0.0085043 T^2 - 0.0000679 T^3
Differentiating with respect to T gives
V'(T) = -0.06426 + 0.0170086 T - 0.0002037 T^2
V"(T) = 0.0170086 - 0.0004074 T

V'(T) = 0 when T = 3.97°C or 79.5°C (3 s.f. - from the quadratic formula), with the latter value being outside of the domain of the given function. At T = 3.97°C V"(T) is positive, so this is a point of minimum volume and therefore maximum density.

Answer 3

The excact maximum density off water is on 4 centigrade
cause water have an unususal changing in volume in 4 and it's volume increases in both ways out, so the density decrease after or before 4!
maximum density is excactly at 4 celcious!

Answer 4

V = 999.87 - 0.06426(T) + 0.0085043(T^2) - 0.0000679(T^3)
dV/dT=0-0.06426+0.0170086T-0.0002037T^2
To get T for maximum density, dV/dT=0
-315.464+83.5T-T^2=0
T^2-83.5T+315.464=0
solve this quadratic equation.

Answer 5

optimal fee/minimum fee: the fee of y coordinate of the vertex of a quadratic equation in basic terms get the by-made of the quadratic equation: f(x) = 12x-x^2 f'(x) = a million(12)(x)^(a million-a million) - 2(x)^(2-a million) 0 = 12 - 2x x = 6 replace it to f(x) = 12x-x^2 = 36 or in basic terms, get the fee of y interior the vertex (x,y) x = -b/2a ; y = (4ac-b^2)/4a y = 4ac -b^2/4a = 36 ---> optimal fee