# Cuboid tank help

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• Nov 28th 2009, 04:55 PM
user_5
Cuboid tank help
A cuboid tank is open at the top and the internal dimensions of its base are x m and 2x m.
The height is h m.
The volume of the tank is V cubic metres and the volume is fixed. Let S mē denote the internal surface area of the tank.

S in terms of x and h
$\displaystyle S=2(2xh)+2x(x)+2(hx)$
$\displaystyle S=2x^2+6xh$

S in terms of V and x
$\displaystyle V=2x * x * h$

$\displaystyle V=2x^2h$

$\displaystyle h=\frac{V}{2x^2}$

$\displaystyle S=2x^2+6xh$

Sub $\displaystyle h=\frac{V}{2x^2}$

$\displaystyle S=2x^2+6x\left(\frac{V}{2x^2}\right)$

$\displaystyle S=2x^2+\frac{3V}{x}$

The maximal domain of S in terms of V and x is $\displaystyle (0,\infty)$

If $\displaystyle 2<x<15$ find the maximum value of S if $\displaystyle V = 1000m^3$

I'm stuck here, can anyone please help?

Here's a visual representation of the cuboid
• Nov 28th 2009, 05:02 PM
skeeter
Quote:

Originally Posted by user_5
A cuboid tank is open at the top and the internal dimensions of its base are x m and 2x m.
The height is h m.
The volume of the tank is V cubic metres and the volume is fixed. Let S mē denote the internal surface area of the tank.

S in terms of x and h
$\displaystyle S=2(2xh)+2x(x)+2(hx)$
$\displaystyle S=2x^2+6xh$

S in terms of V and x
$\displaystyle V=2x * x * h$

$\displaystyle V=2x^2h$

$\displaystyle h=\frac{V}{2x^2}$

$\displaystyle S=2x^2+6xh$

Sub $\displaystyle h=\frac{V}{2x^2}$

$\displaystyle S=2x^2+6x\left(\frac{V}{2x^2}\right)$

$\displaystyle S=2x^2+\frac{3V}{x}$

The maximal domain of S in terms of V and x is $\displaystyle (0,\infty)$

If $\displaystyle 2<x<15$ find the maximum value of S if $\displaystyle V = 1000m^3$

I'm stuck here, can anyone please help?

using calculus, find $\displaystyle \frac{dS}{dx}$ , determine the critical value and confirm that it is a value of x that minimizes S

using technology, graph S(x) in the window 2 < x < 15 and locate the minimum.