Applications of Differentiation

To find the maximum volume possible, we must find two equations in volume and area:

 

V = 5x \times 3x \times h = 15x^2h

 

and

 

A = 2(5x \times 3x) + 2(5x \times h) + 2(3x \times h) = 30x^2 + 16xh = 30000

 

Of course, we only want to differentiate in terms of one variable, so we’ll transform the second equation to

 

h = \dfrac{30000 - 30x^2}{16x}

 

Plugging this back into our equation for V, we have

 

V = \dfrac{15x^2(30000 - 30x^2)}{16x} = 28125x - 28.125x^3

 

We can then differentiate with respect to x to find the maximum volume of the box:

 

\dfrac{dV}{dx} = 28125 - 84.375x^2 = 0, giving x = \sqrt{\dfrac{1000}{3}}

 

We must also verify that this is a maximum, rather than a minimum, so we find \dfrac{d^2V}{dx^2}:

 

\dfrac{d^2V}{dx^2} = - 168.75x = -168.75\sqrt{\dfrac{1000}{3}} < 0, so this is a maximum point.

 

Therefore, we have x = \sqrt{\dfrac{1000}{3}} and h = \dfrac{20000}{16\sqrt{\dfrac{1000}{3}}}, giving V_{max} = \dfrac{6250000\sqrt{3}}{\sqrt{1000}}.

h = 1000\left( \dfrac{t^2}{40} - \dfrac{t^3}{600}\right) gives

 

\dfrac{dh}{dt} = 1000 \left( \dfrac{t}{20} - \dfrac{t^2}{200}\right)

 

and

 

\dfrac{d^2h}{dt^2} = 1000 \left( \dfrac{1}{20} - \dfrac{t}{100}\right)

 

Setting \dfrac{dh}{dt} = 0 gives 1000\left( \dfrac{t}{20}\left( 1 - \dfrac{t}{10}\right) \right) = 0, meaning that t = 0 or t = 10.

 

Using t=10 gives h=833\text{ m}.

Volume of bath, V:

 

V = \dfrac{1}{2}(2.5 + 1.5) \times 0.5 \times 1 = 1\text{ m}^3 = 1 \times 10^6 \text{ cm}^3

 

Then, we must find t such that \dfrac{1}{2}t^2 + 4t - 1000000 = 0.

 

Using the quadratic formula, we have

 

t = \dfrac{-4 \pm \sqrt{16 - \left( 4 \times \dfrac{1}{2} \times -1000000 \right)}}{1} = -1418.22, 1410.22\text{ s}

 

So, we know that the bath takes approximately 23.5\text{ minutes} to fill.

 

When t = 1410.22, \dfrac{dV_{filled}}{dt} = t + 4, so the water is filling the bath at a rate of 1414.22\text{ cm}^3\text{ s}^{-1}.