If the volume of a cylindrical block is equal to 800cm^3 prove that the total surface area is equal to 2(pi)x^2 + (1600)/x, where x cm is the radius of the base. hence obtain the value x which makes the surface area a minimum.
To begin, let's calculate the volume of the cylindrical block. The volume of a cylinder is given by the formula V = πr^2h, where r is the radius of the base and h is the height.
In this case, we are given that the volume is 800 cm^3, so we can write the equation as:
800 = πr^2h
Now, let's express the height 'h' in terms of the radius 'r'. We can do this by rearranging the formula for volume:
h = 800 / (πr^2)
Next, let's calculate the total surface area of the cylindrical block. The surface area of a cylinder is given by the formula A = 2πr^2 + 2πrh.
Substituting the expression for 'h' we derived earlier, we can write the equation as:
A = 2πr^2 + 2πr(800 / πr^2)
Simplifying this equation, we get:
A = 2πr^2 + 1600 / r
Now, we need to find the value of 'x' that minimizes the surface area 'A'. To do this, we can take the derivative of 'A' with respect to 'r' and set it equal to zero. This will give us the critical point(s) where the surface area is either a minimum or maximum.
Let's differentiate 'A' with respect to 'r':
dA/dr = 4πr - 1600 / r^2
Setting this derivative equal to zero:
4πr - 1600 / r^2 = 0
Multiplying through by r^2 to remove the fraction:
4πr^3 - 1600 = 0
Simplifying this equation further:
r^3 = 400 / π
Taking the cube root of both sides:
r = (400 / π)^(1/3)
Now we have obtained the value of 'x' (the radius) that makes the surface area a minimum, which is:
x = (400 / π)^(1/3)
Substituting this value back into the equation for the surface area:
A = 2πx^2 + 1600 / x
we can simplify the equation to:
A = 2π((400 / π)^(2/3)) + 1600 / ((400 / π)^(1/3))
Thus, we have proven that the total surface area of the cylindrical block is equal to 2πx^2 + (1600) / x, where x cm is the radius of the base. We have also obtained the value of x which makes the surface area a minimum.