Maximal Aquarium

Let the width of the square base be $w$ and the height of the aquarium by $h$ . Then its volume is given by $V = w^2h$ and its surface area

$\begin{aligned} 4wh + w^2 & = 400 & \small \color{#3D99F6} \text{Multiply both sides by }w \\ 4{\color{#3D99F6}w^2h} + w^3 & = 400w & \small \color{#3D99F6} \text{Note that }V=w^2h \\ 4{\color{#3D99F6}V} + w^3 & = 400w & \small \color{#3D99F6} \text{Differentiate both sides w.r.t. }w. \\ 4\dfrac {dV}{dw} + 3w^2 & = 400 \end{aligned}$

We note that $\dfrac {dV}{dw} = 0$ , when $w = \dfrac {20}{\sqrt 3}$ and $\dfrac {d^2 V}{dw} < 0$ . This means $V$ is maximum when $w=\dfrac {20}{\sqrt 3}$ . From

$\begin{aligned} 4wh + w^2 & = 400 \\ 4\left(\frac {20}{\sqrt 3}\right)h + \left(\frac {20}{\sqrt 3}\right)^2 & = 400 \\ \frac h{\sqrt 3} + \frac 53 & = 5 \\ \implies h & = \boxed{5\sqrt 3 - \dfrac {5\sqrt 3}3} \end{aligned}$

Let the width be x and height be h (I'd like to see a solution with h=x), SA = x^2 + 4xh. We already know that the surface area is 400 cm^2 so then we shall solve for h

• 400 = x^2+4xh

• $\frac{400-x^2}{4x}$ = h

Now that we have solved for the height in terms of the width it is time to use the volume equation.

• V = x^2*h

• V = x^2*( $\frac{400-x^2}{4x}$ )

• V = $\frac{400x-x^3}{4}$

Now we solve for the derivatives

• V' = 1/4(400-3x^2)

• V" = 1/4(-6x)

We see that the volume is concave down for all positive x Now lets solve for the critical points of the volume

• V' = (400-3x^2)/4 = 0

• 400-3x^2 = 0

• 400=3x^2

• x^2=400/3

So now we know that x is the positive or negative (for our purposes it is positive) square root of 400/3

If we use the height equation

• $\frac{400-x^2}{4x}$ = h

400-400/3 / 80/sqrt(3)

400sqrt(3) - 400sqrt(3)/3 / 80

h=5sqrt(3) - 5sqrt(3)/3