a) A farmer has 600 metres of fencing material with which to enclose a rectangular pen adjacent to a long existing wall.He will use the wall for one side of the fence and the available fencing material for the remaining three sides. What is the maximum area that can be enclosed in this way? (7 marks) b) Given that f(x) is continuous everywhere. Find the values of a and b ,if f(x)= ) -5&if&xlt 1 ax^2-b&if&1leqslant xlt 1 20x&if&xgt 2 (7 marks) c) Find the equation of the line tangent and line normal to the curve given below at the given point xy=6e^2x-3y at (3,2) (6 marks)

Question

a) A farmer has 600 metres of fencing material with which to enclose a rectangular pen adjacent to a long existing wall.He will use the wall for one side of the fence and the available fencing material for the remaining three sides. What is the maximum area that can be enclosed in this way? (7 marks) b) Given that f(x) is continuous everywhere. Find the values of a and b ,if f(x)= ) -5&if&xlt 1 ax^2-b&if&1leqslant xlt 1 20x&if&xgt 2 (7 marks) c) Find the equation of the line tangent and line normal to the curve given below at the given point xy=6e^2x-3y at (3,2) (6 marks)

Answer 1

a) To maximize the area enclosed by the rectangular pen, we need to find the optimal dimensions of the rectangle. Let's denote the length of the rectangle as 'L' and the width as 'W'. Since the farmer is using the existing wall as one side of the fence, the total length of the fencing material will be used for the other three sides. Therefore, we have the equation:L + 2W = 600We want to maximize the area of the rectangle, which is given by:Area = L * WTo find the maximum area, we can express L in terms of W using the equation above:L = 600 - 2WSubstituting this into the area equation, we get:Area = (600 - 2W) * WTo find the maximum area, we can take the derivative of the area equation with respect to W and set it equal to zero:d(Area)/dW = 600 - 4W = 0Solving for W, we get:W = 150Substituting this value back into the equation for L, we get:L = 600 - 2(150) = 300Therefore, the maximum area that can be enclosed is:Area = L * W = 300 * 150 = 45,000 square meters.b) To find the values of a and b, we need to ensure the continuity of the function f(x) at the given points. The function is defined as:f(x) = -5 if x 2For the function to be continuous at x = 1, the value of f(x) from the left side (x 2). Therefore, we have:ax^2 - b = 20xSubstituting x = 2 into this equation, we get:a(2)^2 - b = 20(2)Simplifying this equation, we get:4a - b = 40Now we have a system of two equations with two unknowns:-5 = a - b4a - b = 40We can solve this system of equations to find the values of a and b. Solving for a, we get:a = 45Substituting this value back into the first equation, we get:-5 = 45 - bSolving for b, we get:b = 50Therefore, the values of a and b are a = 45 and b = 50.c) To find the equation of the tangent line to the curve xy = 6e^(2x - 3y) at the point (3, 2), we need to find the derivative of the curve with respect to x. We can use implicit differentiation to find the derivative.Differentiating both sides of the equation with respect to x, we get:y + x(dy/dx) = 2e^(2x - 3y)(1 - 3(dy/dx))Now, we can substitute the point (3, 2) into this equation to find the slope of the tangent line:2 + 3(dy/dx) = 2e^(2(3) - 3(2))(1 - 3(dy/dx))Simplifying this equation, we get:3(dy/dx) + 2 = 2e^(6 - 6)(1 - 3(dy/dx))3(dy/dx) + 2 = 2(1 - 3(dy/dx))3(dy/dx) + 2 = 2 - 6(dy/dx)9(dy/dx) = -4(dy/dx) = -4/9Therefore, the slope of the tangent line at the point (3, 2) is -4/9.To find the equation of the tangent line, we can use the point-slope form of a linear equation:y - y1 = m(x - x1)Substituting the point (3, 2) and the slope -4/9 into this equation, we get:y - 2 = (-4/9)(x - 3)Simplifying this equation, we get:y = (-4/9)x + 14/3Therefore, the equation of the tangent line

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