Answer to Question #280919 in Calculus for Kelvin lee Boacon

Question

3. An open rectangular box w/ square ends to hold 6400 cu ft, is to be built at a cost of

Php 75.00 per sq ft. for the base and

Php 25.00 /sq ft for the sides. Find the most economical dimensions.

Answer 1

Solution:

Let the dimensions be $x,x,y$, since it has square ends.

Volume of rectangular box $=6400\ ft^3$

$\Rightarrow x.x.y=6400 \\ \Rightarrow x^2y=6400 \\ \Rightarrow y=\dfrac{6400}{x^2}\ ...(i)$

Total surface area $=S=x.x+2(xy+yx)=x^2+4xy$

So, total cost $=C=(75\times x^2)+(4xy)\times25$

$\Rightarrow C=75 x^2+(100x)(\dfrac{6400}{x^2}) \ [Using\ (i)] \\=75x^2+640000(\dfrac1x)$

On differentiating w.r.t $x$ ,

$C'=150x-\dfrac{640000}{x^2}$

Now put C'=0

$\Rightarrow 150x-\dfrac{640000}{x^2}=0 \\ \Rightarrow 150x^3=640000 \\ \Rightarrow x^3=\dfrac{128000}{3} \\ \Rightarrow x\approx 34.94$

Again differentiating C' w.r.t x,

$C''=150+\dfrac{128000}{x^3}>0, for\ x=34.94$

Thus, minima exists.

Put the value of x in (i).

$y=\dfrac{6400}{34.94^2}\approx5.24$

Thus, most economical dimensions are 34.94 ft, 34.94 ft and 5.24 ft.