Answer to Question #145632 in Calculus for Abdulsalam Aminu Bandam

Question

The volume of water entering a tank from a pump is related by the formula v=ah^3+bh^2+ch+d where is the height of the tank, a, b, c and d are constants.
Fine the condition for minimum and maximum water storage.

Answer 1

$V=ah^3+bh+ch+d$

at Maximum and Minimum Volume, $\frac{dV}{dh}=0$

getting the derivative of V w.r.t to h we obtain

$\frac{dV}{dh}=3ah^2+2bh+c$

this implies that, $3ah^2+2bh+c=0$

solving for h using that maximizes/minimizes Volume using completing square we get:

$3ah^2+2bh=-c$

dividing by 3a, $h^2+\frac{2b}{3a}h=\frac{-c}{3a}$

completing the square, $h^2+\frac{2b}{3a}h+(\frac{b}{3a})^2=\frac{-c}{3a}+\frac{b^2}{9a^2}$

$(h+\frac{b}{3a})^2= \frac{b^2-3ac}{9a^2}$

$(h+\frac{b}{3a})=\sqrt\frac{b^2-3ac}{9a^2}$

$h1= \frac{b}{3a}+\frac{\sqrt {(b^2-3ac)}}{3a}$

$h1= \frac{-b+\sqrt{(b^2-3ac)}}{3a}$

$h2= \frac{b}{3a}-\frac{\sqrt{(b^2-3ac)}}{3a}$

$h2= \frac{-b-\sqrt{(b^2-3ac)}}{3a}$

therefore, the volume will be maximum/minimum when

$h1= \frac{-b+\sqrt{(b^2-3ac)}}{3a}$ and

$h2= \frac{-b-\sqrt{(b^2-3ac)}}{3a}$ (condition)

At Maximum

$\frac{dV}{dh}<0$ at H1 and H2

At minimum

$\frac{dV}{dh}>0$ at H1 and H2