At an Oregon fiber-manufacturing facility, an analyst estimates that the weekly number of pounds of acetate fibers that can be produced is given by the function: \[ z=f(x, y)=14500 x+3000 y+5 x^{2} y-7 x^{3} \] Where: \( z= \) the weekly \# of pounds of acetate fiber \( x= \) the \# of skilled workers at the plant \( y= \) the \# of unskilled workers at the plant Determine the following: A) The weekly number of pounds of fiber that can be produced with 13 skilled workers and 32 unskilled workers. Answer = \( \square \) pounds B) Find an expression \( \left(f_{x}\right) \) for the rate of change of output with respect to the number of skilled workers. Answer \( =f_{x}= \) \( \square \) C) Find an expression \( \left(f_{y}\right) \) for the rate of change of output with respect to the number of unskilled workers. Answer \( =f_{y}= \) \( \square \) D) Find the rate of change of output with respect to skilled workers when 13 skilled workers and 32 unskilled workers are employed. (Your answer will be a number.) - Answer = \( \square \) weekly pounds per skilled worker E) Find the rate of change of output with respect to unskilled workers when 13 skilled workers and 32 unskilled workers are employed. (Your answer will be a number.)

To solve the problem step by step: ### A) Weekly pounds of fiber produced with 13 skilled workers and 32 unskilled workers. We need to substitute \( x = 13 \) and \( y = 32 \) into the function \( z = f(x, y) \). \[ z = 14500(13) + 3000(32) + 5(13^2)(32) - 7(13^3) \] Calculating each term: - \( 14500(13) = 188500 \) - \( 3000(32) = 96000 \) - \( 5(169)(32) = 27040 \) (since \( 13^2 = 169 \)) - \( 7(2197) = 15379 \) (since \( 13^3 = 2197 \)) Now putting it all together: \[ z = 188500 + 96000 + 27040 - 15379 = 280161 \] **Answer**: 280161 pounds ### B) Expression \( f_{x} \) for the rate of change with respect to skilled workers To find \( f_{x} \), we need to take the partial derivative of the function with respect to \( x \): \[ f(x, y) = 14500x + 3000y + 5x^2y - 7x^3 \] Taking the derivative: \[ f_{x} = \frac{\partial}{\partial x}(14500x) + \frac{\partial}{\partial x}(3000y) + \frac{\partial}{\partial x}(5x^2y) - \frac{\partial}{\partial x}(7x^3) \] Calculating: \[ f_{x} = 14500 + 0 + 10xy - 21x^2 \] **Answer**: \( f_{x} = 14500 + 10xy - 21x^2 \) ### C) Expression \( f_{y} \) for the rate of change with respect to unskilled workers Now, we take the partial derivative of the function with respect to \( y \): \[ f_{y} = \frac{\partial}{\partial y}(14500x) + \frac{\partial}{\partial y}(3000y) + \frac{\partial}{\partial y}(5x^2y) - \frac{\partial}{\partial y}(7x^3) \] Calculating: \[ f_{y} = 0 + 3000 + 5x^2 + 0 \] **Answer**: \( f_{y} = 3000 + 5x^2 \) ### D) Rate of change of output with respect to skilled workers at \( x=13 \), \( y=32 \) Substitute \( x=13 \) and \( y=32 \) into the expression for \( f_{x} \): \[ f_{x} = 14500 + 10(13)(32) - 21(13^2) \] Calculating each term: - \( 10(13)(32) = 4160 \) - \( 21(169) = 3549 \) Now putting it all together: \[ f_{x} = 14500 + 4160 - 3549 = 17211 \] **Answer**: 17211 weekly pounds per skilled worker ### E) Rate of change of output with respect to unskilled workers at \( x=13 \), \( y=32 \) Substitute \( x=13 \) into the expression for \( f_{y} \): \[ f_{y} = 3000 + 5(13^2) \] Calculating: \[ 5(169) = 845 \] So, \[ f_{y} = 3000 + 845 = 3845 \] **Answer**: 3845 weekly pounds per unskilled worker