This mathematics problem involves applying core mathematical principles and formulas. Below you will find a complete step-by-step solution with detailed explanations for each step, helping you understand not just the answer but the method behind it.
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Here are the solutions to your questions. QUESTION 10 (Cylinder Volume) 10.1 Step 1: Identify the dimensions of the cylinder based on the given volume formula. The volume of a cylinder is given by V = r^2 h. The given formula is V = (25x^2)/(4) - (x^3)/(4) = (x^2)/(4)(25-x). Comparing the two forms, we can infer the relationships between r, h, and x. If we assume that the circumference of the cylinder's base is x, then: 2 r = x r = (x)/(2). If we assume the height of the cylinder is h = 25-x. Step 2: Substitute these assumed dimensions into the volume formula. V = r^2 h V = ((x)/(2))^2 (25-x) V = ((x^2)/(4^2)) (25-x) V = (x^2)/(4) (25-x) V = (25x^2)/(4) - (x^3)/(4) This matches the given formula. (This derivation assumes that the circumference of the cylinder's base is x and its height is 25-x). 10.2 Step 1: Differentiate the volume function with respect to x. The volume function is V(x) = (25x^2)/(4) - (x^3)/(4). To find the maximum volume, we need to find the derivative V'(x) and set it to zero. V'(x) = (d)/(dx)((25x^2)/(4) - (x^3)/(4)) V'(x) = (25)/(4)(2x) - (1)/(4)(3x^2) V'(x) = (50x)/(4) - (3x^2)/(4) Step 2: Set the derivative to zero and solve for x. (50x)/(4) - (3x^2)/(4) = 0 Multiply by 4: 50x - 3x^2 = 0 Factor out x: x(50 - 3x) = 0 This gives two possible solutions: x=0 or 50-3x=0. x=0 would result in zero volume, which is a minimum. For 50-3x=0: 3x = 50 x = (50)/(3) Step 3: Verify that this value of x corresponds to a maximum. We can use the second derivative test. V''(x) = (d)/(dx)((50x)/(4) - (3x^2)/(4)) V''(x) = (50)/(4) - (6x)/(4) Substitute x = (50)/(3): V''((50)/(3)) = (50)/(4) - (6(50)/(3))4 V''((50)/(3)) = (50)/(4) - (100)/(4) V''((50)/(3)) = -(50)/(4) Since V''((50)/(3)) < 0, the volume is maximized at x = (50)/(3). The value of x that will maximise the volume of the cylinder is (50)/(3) units. --- QUESTION 10 (Parallelogram in Triangle) 10.1 Step 1: Use the properties of similar triangles. Given that DECF is a parallelogram, DE FC. Since F is on BC, this means DE BC. Because DE BC, ADE is similar to ABC. AG is the perpendicular height from A to BC, and H is the intersection of AG and DE. Due to the similarity of ADE and ABC, the ratio of their corresponding heights from vertex A is equal to the ratio of their corresponding bases. So, (AH)/(AG) = (DE)/(BC). Step 2: Express DE and BC in terms of the given ratio. We are given BF : FC = 2 : 3. Let BF = 2k and FC = 3k. Then BC = BF + FC = 2k + 3k = 5k. Since DECF is a parallelogram, DE = FC = 3k. Now substitute these into the ratio from Step 1: (AH)/(AG) = (3k)/(5k) = (3)/(5) This means AH = (3)/(5) AG. Step 3: Determine the ratio AH : HG. We know that AG = AH + HG. Substitute AH = (3)/(5) AG: (3)/(5) AG + HG = AG HG = AG - (3)/(5) AG HG = (2)/(5) AG Now find the ratio AH : HG: (AH)/(HG) = (3)/(5) AG(2)/(5) AG = (3)/(2) The ratio AH : HG is 3 : 2. 10.2 Step 1: Express the dimensions of the parallelogram in terms of t. Given AG = t units. From 10.1, AH = (3)/(5) AG = (3)/(5) t. And HG = (2)/(5) AG = (2)/(5) t. This is the height of the parallelogram. We are given BC = (5-t) units. Since FC = (3)/(5) BC (from BF:FC=2:3), the base of the parallelogram FC = DE = (3)/(5)(5-t). Step 2: Write the area of the parallelogram as a function of t. Area of parallelogram A(t) = base × height A(t) = FC × HG A(t) = ((3)/(5)(5-t)) × ((2)/(5)t) A(t) = (6)/(25) (5-t)t A(t) = (6)/(25) (5t - t^2) Step 3: Differentiate the area function and set it to zero to find the maximum. A'(t) = (d)/(dt)((6)/(25) (5t - t^2)) A'(t) = (6)/(25) (5 - 2t) Set A'(t) = 0: (6)/(25) (5 - 2t) = 0 5 - 2t = 0 2t = 5 t = (5)/(2) Step 4: Verify that this value of t corresponds to a maximum. We can use the second derivative test. A''(t) = (d)/(dt)((6)/(25) (5 - 2t)) A''(t) = (6)/(25) (-2) A''(t) = -(12)/(25) Since A''(t) < 0, the area is maximized at t = (5)/(2). The value of t for which the area of the parallelogram is a maximum is (5)/(2) units. --- QUESTION 9 (Distance Minimization) Step 1: Define the coordinates of Benny and the car. Benny's position is B(0, 3). The car's position P(x, y) is on the road described by y = x^2 + 2. So, P(x, x^2+2). The condition x 0 is given. Step 2: Write the distance formula between Benny and the car. The distance D between two points (x_1, y_1) and (x_2, y_2) is D = sqrt((x_2-x_1)^2 + (y_2-y_1)^2). Let's minimize the square of the distance, D^2, to avoid the square root. D^2 = (x - 0)^2 + ((x^2+2) - 3)^2 D^2 = x^2 + (x^2 - 1)^2 Step 3: Expand and simplify the expression for D^2. Let f(x) = D^2. f(x) = x^2 + (x^4 - 2x^2 + 1) f(x) = x^4 - x^2 + 1 Step 4: Differentiate f(x) with respect to x and set it to zero. f'(x) = (d)/(dx)(x^4 - x^2 + 1) f'(x) = 4x^3 - 2x Set f'(x) = 0: 4x^3 - 2x = 0 Factor out 2x: 2x(2x^2 - 1) = 0 This gives two possible solutions for x: 2x = 0 x = 0 2x^2 - 1 = 0 2x^2 = 1 x^2 = (1)/(2) x = ± (1)/(sqrt(2)) Step 5: Consider the valid values of x and find the minimum D^2. Since x 0, we consider x=0 and x = (1)/(sqrt(2)). Evaluate f(x) at these points: For x=0: f(0) = (0)^4 - (0)^2 + 1 = 1 For x = (1)/(sqrt(2)): f((1)/(sqrt(2))) = ((1)/(sqrt(2)))^4 - ((1)/(sqrt(2)))^2 + 1 f((1)/(2)) = (1)/(4) - (1)/(2) + 1 f((1)/(2)) = (1)/(4) - (2)/(4) + (4)/(4) = (3)/(4) The minimum value of D^2 is (3)/(4), which occurs at x = (1)/(sqrt(2)). Step 6: Calculate the minimum distance D. D = sqrt(D^2) = sqrt((3)/(4)) = sqrt(3)sqrt(4) = sqrt(3)2 The distance between Benny and the car when the car is closest to Benny is sqrt(3)2 units. Got more? Send 'em!