A tank with a rectangular base and rectangular sides, open at the top is to be constructed so that its depth is 2 m and volume is 8 m3. If the building of tank costs Rs. 70 per square metre for the base and Rs. 45 per square metre for the sides, what is the cost of least expensive tank?

 

 

 

 
 
 
 
 

Answers (1)

Givenh = 2

            Volume = 8 

            \begin{matrix} lbh = V\\ lbh = 8\\ lb = 4 \end{matrix}\qquad \begin{matrix} \text{length} = x \\ \text{breath} = y \end{matrix}

               xy = 4 \Rightarrow y = \frac{x}{4}

Let C be the cost of the tank then,

            C = 70xy +45(2\times 2x + (2\times 2y))

                  = 70xy +180 x + 180 y

                  = 70x\frac{4}{x} +180 x + 180 \frac{4}{x}

                  =280 +180 x + \frac{720}{x}\qquad - (i)

Now, differentiating both sides with respect to x, we get

\frac{\text{d} C}{\text{d}x} = 180 -\frac{720}{x^2}

For maxima & minima \Rightarrow \frac{\text{d} C}{\text{d}x} = 0

180 -\frac{720}{x^2} = 0 \\\Rightarrow 180 = \frac{720}{x^2} \\\Rightarrow x^2 = \frac{720}{180} \\\Rightarrow x^2 = 4\Rightarrow x =\pm 2

Again, differentiating both sides w.r.t x, we get

\frac{\text d^2 C}{\text d x^2} = \frac{1440}{x^3}

\left .\frac{\text d^2 C}{\text d x^2}\right |_{\text {at }x = 2} = \frac{1440}{8} = 180 > 0

\therefore When x =2, the cost of the tank is minimum

Substituting the value of x in equation (i), we get

            C = 280 + 180\times 2 + \frac{720}{2}

                  = 280 + 180\times 2 + 360

                  = 280 +360 + 360

                  = 1000

Hence, the cost of the least expensive tank is Rs. 1000.

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