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# A tank with a rectangular base and rectangular sides open at the top is to be contrasted so that its depth is 2m and volume is $8{m^3}$. If building a tank costs R.s. 70 per square metre for the base and R.s. 45 per square metre for sides, what is the cost of the least expensive tank?

Last updated date: 13th Jun 2024
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Hint: First, we will proceed by calculating the volume of the tank and area of the base. After that, we will find the area of sides and then use differentiation to find the length of sides. Thus, we will find the least cost of a tank by substituting the value of length and breadth.

Complete step-by-step solution:
Given that the depth of the tank is 2m and the volume of the tank is $8{m^3}$.
Let the length of the tank be $x$and the breadth of the tank be $y$.
Also, we know that,
Volume of tank $= l \times b \times h$
Substitute the values,
$\Rightarrow 8 = x \times y \times 2$
Divide both sides by 2,
$\Rightarrow xy = 4$
Find the value of $y$ in terms of $x$,
$\Rightarrow y = \dfrac{4}{x}$...................….. (1)
Also, we are given the cost of building the tank Rs. 70 per square metre for the base. Then,
Area of the rectangular base $= l \times b$
Substitute the values,
$\Rightarrow$Area of the rectangular base $= xy$
So, the cost of the rectangular base will be,
$\Rightarrow {C_b} = 70xy$
Substitute the values from equation (1),
$\Rightarrow {C_b} = 70x \times \dfrac{4}{x}$
Simplify the terms,
$\Rightarrow {C_b} = 280$.................….. (2)
Also, we are given that cost is Rs. 45 per square metre for rectangular sides.
Area of the rectangular sides $= 2\left( {lh + bh} \right)$
Substitute the values,
$\Rightarrow$Area of the rectangular sides $= 2\left( {2x + 2y} \right)$
Take 2 common from both terms in the bracket,
$\Rightarrow$Area of the rectangular sides $= 4\left( {x + y} \right)$
So, the cost of rectangular sides will be,
$\Rightarrow {C_s} = 45 \times 4\left( {x + y} \right)$
Multiply the terms,
$\Rightarrow {C_s} = 180\left( {x + y} \right)$
Substitute the value from equation (1),
$\Rightarrow {C_s} = 180\left( {x + \dfrac{4}{x}} \right)$....................….. (3)
Let the total cost of the tank be $C$. Then,
$\Rightarrow C\left( x \right) = {C_b} + {C_s}$
Substitute the values from equation (2) and (3),
$\Rightarrow C\left( x \right) = 280 + 180\left( {x + \dfrac{4}{x}} \right)$..................….. (4)
Now differentiate the above equation w.r.t. $x$ to find the minimum cost,
$\Rightarrow C'\left( x \right) = \dfrac{d}{{dx}}\left( {280 + 180\left( {x + \dfrac{4}{x}} \right)} \right)$
Using the property $\dfrac{d}{{dx}}\left( {f + g} \right) = \dfrac{{df}}{{dx}} + \dfrac{{dg}}{{dx}}$, we get,
$\Rightarrow C'\left( x \right) = 0 + 180\left( {1 + \dfrac{4}{{{x^2}}} \times \left( { - 1} \right)} \right)$
Simplify the terms,
$\Rightarrow C'\left( x \right) = 180\left( {1 - \dfrac{4}{{{x^2}}}} \right)$
Now equate it with 0 to get the minimum cost,
$\Rightarrow 180\left( {1 - \dfrac{4}{{{x^2}}}} \right) = 0$
Divide both sides by 180,
$\Rightarrow 1 - \dfrac{4}{{{x^2}}} = 0$
Move variable part on another side,
$\Rightarrow 1 = \dfrac{4}{{{x^2}}}$
Cross-multiply the terms,
$\Rightarrow {x^2} = 4$
Take the square root on both sides,
$\Rightarrow x = \pm 2$
Since, the length cannot be negative,
$\Rightarrow x = 2$
Again differentiate wrt to $x$ and substitute the value to check whether it is the point of minima.
$\Rightarrow C''\left( x \right) = \dfrac{d}{{dx}}\left( {180\left( {1 - \dfrac{4}{{{x^2}}}} \right)} \right)$
Using the property $\dfrac{d}{{dx}}\left( {ax} \right) = a\dfrac{{dx}}{{dx}}$, we get,
$\Rightarrow C''\left( x \right) = 180\dfrac{d}{{dx}}\left( {1 - \dfrac{4}{{{x^2}}}} \right)$
Using the property $\dfrac{d}{{dx}}\left( {f + g} \right) = \dfrac{{df}}{{dx}} + \dfrac{{dg}}{{dx}}$, we get,
$\Rightarrow C''\left( x \right) = 180\left( {0 - \dfrac{4}{{{x^3}}} \times - 2} \right)$
Simplify the term,
$\Rightarrow C''\left( x \right) = 180 \times \dfrac{8}{{{x^3}}}$
Substitute the value of $x$,
$\Rightarrow C''\left( 2 \right) = 180 \times \dfrac{8}{{{2^3}}}$
Simplify the terms,
$\Rightarrow C''\left( 2 \right) = 180 > 0$
So, $C\left( x \right)$ is minimum at $x = 2$.
Substitute the value of $x$ in equation (4) to get the minimum cost,
$\Rightarrow C\left( 2 \right) = 280 + 180\left( {2 + \dfrac{4}{2}} \right)$
Simplify the term,
$\Rightarrow C\left( 2 \right) = 280 + 180\left( {2 + 2} \right)$
$\Rightarrow C\left( 2 \right) = 280 + 180 \times 4$
$\Rightarrow C\left( 2 \right) = 280 + 720$
$\therefore C\left( 2 \right) = 1000$