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Find the maximum value of f(x, y) = x^3y^8 for x, y ≥ 0 on the unit circle. fmax= ____.
Find the maximum value of f(x, y) = x^3y^8 for x, y ≥ 0 on the unit circle. fmax= ____.
Answer:fmax = (12288√33)/1771561 ≈ 0.03984575513Step-by-step explanation:You want themaximum valueof theexpression f(x, y) = x³y⁸in thefirst quadrantof theunit circle.SubstitutionThe unit circle has the equation ...x² +y² = 1This lets us write an expression for y²:y² = 1 -x²Using this in the definition of f(x, y), we can write ...f(x) = x³(1 -x²)⁴MaximumThe derivative of this function of x is ...f'(x) = 3x²(1 -x²)⁴ +4x³(1 -x²)³(-2x) = (x²(1 -x²)³)·(3(1 -x²) -8x²)This is zero at the critical points:x = 0, x = 1, and 3 -11x² = 0, or x = √(3/11)Then the maximum value of f is ...f(√(3/11)) = (√(3/11))³(1 -3/11)⁴ = (√(3/11))(3/11)(8/11)⁴fmax = 3·8⁴√33/11⁶fmax = (12288√33)/1771561 ≈ 0.03984575513__Additional commentThe attached graph shows the curve f(x, y) and the unit circle constraint. The value of f(x, y) that makes the curves tangent is the one shown above.<95141404393>...