Question

A rectangular tank that is 8788 ft cubed with a square base and open top is to be constructed of sheet steel of a given thickness. find the dimensions of the tank with minimum weight.

Answer 1

Form minimum weight, the surface area must be minimum.
Let the height be h and the lengths be x
the volume will be:
8788=hx^2
hence
h=8788/x^2
A=x^2+4xh
A=x^2+35142/x
dA/dx=2x-35152/x^2
thus
when dA/dx=0
2x-35152/x^2=0
2x=35152/x^2
hence
2x^3=35152
x^3=17576
x=26 ft
hence
h=8788/x^2=8788/26^2=13 ft
Hence the dimensions are 26 ft by 26 ft by 13 ft

Answer 2

The dimensions of the rectangular tank are $26\;\rm{ft}\times26\;\rm{ft}\times13\;\rm{ft}$.

Step-by-step explanation:

Given: A rectangular tank that is $8788\;\rm{ft^3$ with a square base and open top is to be constructed of sheet steel of a given thickness.

According to question:

For the value of minimum weight, the surface area sholud be minimum.

Let the height be $h$ and the length of the tank be $x$.

So, the volume of tank will be formulated as: $V=lbh$

$hx^2=8788$

$h=\frac{8788}{x^2}$

Now,

$A=x^2+4xh\\A=x^2+\frac{35142}{x}$

Differentiating the area w.r.t $x$ we get:  

$\frac{dA}{dx}=2x-35152/x^2$

For surface area to be minimum then  

$2x-35152/x^2=0$

$2x=35152/x^2$

 

$h=\frac{8788}{26^2} =\frac{8788}{676}=13\;\rm{ft}$

Hence, the dimensions of tank are $26\;\rm{ft}\times26\;\rm{ft}\times13\;\rm{ft}$.

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