A cylindrical water tank has a fixed surface area of A0.
. Find an expression for the maximum volume that such a water tank can take.

Answers

Answer 1

(i) The maximum volume of a cylindrical water tank with fixed surface area A₀ is 0, occurring when the tank is empty. (ii) The indefinite integral of F(x) = 1/(x²(3x - 1)) is F(x) = -ln|x| + 1/x - 3ln|3x - 1| + C.

(i) To find the expression for the maximum volume of a cylindrical water tank with a fixed surface area of A₀ m², we need to consider the relationship between the surface area and the volume of a cylinder.

The surface area (A) of a cylinder is given by the formula:

A = 2πrh + πr²,

where r is the radius of the base and h is the height of the cylinder.

Since the surface area is fixed at A₀, we can express the radius in terms of the height using the equation

A₀ = 2πrh + πr².

Solving this equation for r, we get:

r = (A₀ - 2πrh) / (πh).

Now, the volume (V) of a cylinder is given by the formula:

V = πr²h.

Substituting the expression for r, we can write the volume as:

V = π((A₀ - 2πrh) / (πh))²h

= π(A₀ - 2πrh)² / (π²h)

= (A₀ - 2πrh)² / (πh).

To find the maximum volume, we need to maximize this expression with respect to the height (h). Taking the derivative with respect to h and setting it equal to zero, we can find the critical point for the maximum volume.

dV/dh = 0,

0 = d/dh ((A₀ - 2πrh)² / (πh))

= -2πr(A₀ - 2πrh) / (πh)² + (A₀ - 2πrh)(-2πr) / (πh)³

= -2πr(A₀ - 2πrh) / (πh)² - 2πr(A₀ - 2πrh) / (πh)³.

Simplifying, we have:

0 = -2πr(A₀ - 2πrh)[h + 1] / (πh)³.

Since r ≠ 0 (otherwise, the volume would be zero), we can cancel the r terms:

0 = (A₀ - 2πrh)(h + 1) / h³.

Solving for h, we get:

(A₀ - 2πrh)(h + 1) = 0.

This equation has two solutions: A₀ - 2πrh = 0 (which means the height is zero) or h + 1 = 0 (which means the height is -1, but since height cannot be negative, we ignore this solution).

Therefore, the maximum volume occurs when the height is zero, which means the water tank is empty. The expression for the maximum volume is V = 0.

(ii) To find the indefinite integral of F(x) = ∫(1 / (x²(3x - 1))) dx:

Let's use partial fraction decomposition to split the integrand into simpler fractions. We write:

1 / (x²(3x - 1)) = A / x + B / x² + C / (3x - 1),

where A, B, and C are constants to be determined.

Multiplying both sides by x²(3x - 1), we get:

1 = A(3x - 1) + Bx(3x - 1) + Cx².

Expanding the right side, we have:

1 = (3A + 3B + C)x² + (-A + B)x - A.

Matching the coefficients of corresponding powers of x, we get the following system of equations:

3A + 3B + C = 0, (-A + B) = 0, -A = 1.

Solving this system of equations, we find:

A = -1, B = -1, C = 3.

Now, we can rewrite the original integral using the partial fraction decomposition

F(x) = ∫ (-1 / x) dx + ∫ (-1 / x²) dx + ∫ (3 / (3x - 1)) dx.

Integrating each term

F(x) = -ln|x| + 1/x - 3ln|3x - 1| + C,

where C is the constant of integration.

Therefore, the indefinite integral of F(x) is given by:

F(x) = -ln|x| + 1/x - 3ln|3x - 1| + C.

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--The given question is incomplete, the complete question is given below " (i) A cylindrical water tank has a fixed surface area of A₀ m². Find an expression for the maximum volume that such a water tank can take. (ii) Find the indefinite integral F(x)=∫ 1dx/(x²(3x−1))."--


Related Questions

Find the general solution to the following differential equations:
16y''-8y'+y=0
y"+y'-2y=0
y"+y'-2y = x^2

Answers

The general solution of the given differential equations are:

y = c₁e^(x/4) + c₂xe^(x/4) (for 16y''-8y'+y=0)

y = c₁e^x + c₂e^(-2x) (for y"+y'-2y=0)

y = c₁e^x + c₂e^(-2x) + (1/2)x

(for y"+y'-2y=x²)

Given differential equations are:

16y''-8y'+y=0

y"+y'-2y=0

y"+y'-2y = x²

To find the general solution to the given differential equations, we will solve these equations one by one.

(i) 16y'' - 8y' + y = 0

The characteristic equation is:

16m² - 8m + 1 = 0

Solving this quadratic equation, we get m = 1/4, 1/4

Hence, the general solution of the given differential equation is:

y = c₁e^(x/4) + c₂xe^(x/4)..................................................(1)

(ii) y" + y' - 2y = 0

The characteristic equation is:

m² + m - 2 = 0

Solving this quadratic equation, we get m = 1, -2

Hence, the general solution of the given differential equation is:

y = c₁e^x + c₂e^(-2x)..................................................(2)

(iii) y" + y' - 2y = x²

The characteristic equation is:

m² + m - 2 = 0

Solving this quadratic equation, we get m = 1, -2.

The complementary function (CF) of this differential equation is:

y = c₁e^x + c₂e^(-2x)..................................................(3)

Now, we will find the particular integral (PI). Let's assume that the PI of the differential equation is of the form:

y = Ax² + Bx + C

Substituting the value of y in the given differential equation, we get:

2A - 4A + 2Ax² + 4Ax - 2Ax² = x²

Equating the coefficients of x², x, and the constant terms on both sides, we get:

2A - 2A = 1,

4A - 4A = 0, and

2A = 0

Solving these equations, we get

A = 1/2,

B = 0, and

C = 0

Hence, the particular integral of the given differential equation is:

y = (1/2)x²..................................................(4)

The general solution of the given differential equation is the sum of CF and PI.

Hence, the general solution is:

y = c₁e^x + c₂e^(-2x) + (1/2)x²..................................................(5)

Conclusion: Therefore, the general solution of the given differential equations are:

y = c₁e^(x/4) + c₂xe^(x/4) (for 16y''-8y'+y=0)

y = c₁e^x + c₂e^(-2x) (for y"+y'-2y=0)

y = c₁e^x + c₂e^(-2x) + (1/2)x

(for y"+y'-2y=x²)

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The particular solution is: y = -1/2 x². The general solution is: y = c1 e^(-2x) + c2 e^(x) - 1/2 x²

The general solution of the given differential equations are:

Given differential equation: 16y'' - 8y' + y = 0

The auxiliary equation is: 16m² - 8m + 1 = 0

On solving the above quadratic equation, we get:

m = 1/4, 1/4

∴ General solution of the given differential equation is:

y = c1 e^(x/4) + c2 x e^(x/4)

Given differential equation: y" + y' - 2y = 0

The auxiliary equation is: m² + m - 2 = 0

On solving the above quadratic equation, we get:

m = -2, 1

∴ General solution of the given differential equation is:

y = c1 e^(-2x) + c2 e^(x)

Given differential equation: y" + y' - 2y = x²

The auxiliary equation is: m² + m - 2 = 0

On solving the above quadratic equation, we get:m = -2, 1

∴ The complementary solution is:y = c1 e^(-2x) + c2 e^(x)

Now we have to find the particular solution, let us assume the particular solution of the given differential equation:

y = ax² + bx + c

We will use the method of undetermined coefficients.

Substituting y in the differential equation:y" + y' - 2y = x²a(2) + 2a + b - 2ax² - 2bx - 2c = x²

Comparing the coefficients of x² on both sides, we get:-2a = 1

∴ a = -1/2

Comparing the coefficients of x on both sides, we get:-2b = 0 ∴ b = 0

Comparing the constant terms on both sides, we get:2c = 0 ∴ c = 0

Thus, the particular solution is: y = -1/2 x²

Now, the general solution is: y = c1 e^(-2x) + c2 e^(x) - 1/2 x²

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