Question
Question: A tank with rectangular box and rectangular sides,open at the top is to be constructed so that its d...
A tank with rectangular box and rectangular sides,open at the top is to be constructed so that its depth is 2m and volume is 8m3. If the building of a tank costs Rs.70 per sq metres for the base and RS.45 per square metre for sides, what is the cost of the least expensive tank?
Solution
Hint: Make use of the formula of volume of a cuboid and solve this.
Complete step-by-step answer:
Let us first construct a figure with the data given
Let us consider the length of the box to be= x metres and
The Breadth of the tank be= y metres
The height of the tank is given as =h metres
Volume of the tank is also given as 8cubic metres
The formula for the volume of the tank=l×b×h
⇒8=2×x×y ⇒4=xy ⇒y=x4−−−−−−−−(i)
Given: Building a tank costs Rs.70 per sq.meter for base
Area of base=l×b
Area of base=xy
Cost of base=70(xy)
Also given cost is Rs.45 per square metre
Area of closed sides=2(hl+hb)
=2(2x+2y)
=4(x+y)
Cost of making sides =45[4(x+y)]=180(x+y)
Let C be the total cost of tank
C(x)=Cost of Base+Cost of sides
C(x)=70(xy)+180(x+y)
We got the value of y=x4 from eq (i)
Let’s substitute the value here, so we get
C(x)=70(4)+180(x+x4)
C(x)=280+180(x+4x−1)
We need to minimise the cost of the tank
So, let's find out the minimum of this
So, we will differentiate C(x) with respect to x
So, we get C′(x)=dxd(280+180(x+4x−1))
Putting C’(x)=0,
(x-2)(x+2)=0
So, x=2 or x=-2
Since length cannot be negative , the value of x=2
Finding C′′(x) ,
Differentiating C′(x) with respect to x, we get
C′′(x)=x31440
From this, we get x=2 is a point of minima
Thus, the least cost of construction
C(x)=280+180(2+24)
From this we get 280+720=1000
So, from this we get the least cost of construction to be equal to Rs.1000.
Note: Whenever we have to find the minima, we have to find out the second order derivative and then find out the least cost. One should be careful while differentiating.