Find the critical points of the given function f (x,y)= ye^-(16x^2+9y^2)/288-(16x^2+9y^2)/288 is tthe power

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sciencesolve | Teacher | (Level 3) Educator Emeritus

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You need to remember that you may find the critical points of a two variable function solving the system of equations `f_(x,y) = 0`  and `f_(x,y) = 0`  .

You need to differentiate with respect to x considering y as a constant such that:

`f_x(x,y) = -288y*32x*e^(16x^2+9y^2)`

`f_x(x,y) = -9216xy*e^(16x^2+9y^2)/((288e^(16x^2+9y^2))^2)`

You need to differentiate with respect to y considering x as a constant such that:You need to solve the system of equations such that:

`f_y(x,y) = (288e^(16x^2+9y^2) - 288*18y^2*e^(16x^2+9y^2))/((288e^(16x^2+9y^2))^2)`

`f_y(x,y) = (288e^(16x^2+9y^2)(1 - 18y^2))/((288e^(16x^2+9y^2))^2)`

`f_y(x,y) = (1 - 18y^2)/(288e^(16x^2+9y^2))`

You need to solve the system of equations:

`{(f_y(x y) = 0),(f_y(x y) = 0):}`

`-9216xy*e^(16x^2+9y^2)/((288e^(16x^2+9y^2))^2) = 0`

`xy = 0 => x = 0 or y = 0`

`(1 - 18y^2)/(288e^(16x^2+9y^2)) = 0 => 1 - 18y^2 = 0 => y_1,2 = +-sqrt18/18`

Hence, evaluating the critical points of the function yields `(0,0), (0,-sqrt18/18) , (0,sqrt18/18).`

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