5. Extrema of Multivariable Functions
Example 6 Identify Critical Points
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Extrema of Multivariable Functions
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- Example 7 Maximize Profit
Example 6 Identify Critical PointsUse the Test for Relative Extrema to classify the critical points for f x, y y4 32 y x3 x2 saddle points. as relative maximum, relative minimum, or Solution To find the critical points, we need to compute the first partial derivatives of the function. The first partial derivatives are x y f x, y 3x2 2x f x, y 4 y3 32 Set each partial derivative equal to zero to find the critical points. Let’s start with the partial derivative with respect to x: 0 3x2 2x 0 x 3x 2 x 0 or 3x 2 0 3 x 2 Factor the right side. Set each factor equal to zero to solve for x. Now the partial derivative with respect to y: 0 4 y3 32 0 4 y3 8 0 y3 8 8 y3 2 y Factor the right side. Set each factor equal to zero. Only the factor in parentheses can be zero. Cube root both side to solve for y. 3 Combining these together we get two critical points, 0, 2 and 2 ,2 . We’ll need the second partial derivatives to apply the Test for Relative Extrema to each critical point. fxx x, y 6x 2 f x, y 12 y2 fxy x, y 0 yy Evaluate each second partial derivative at the critical points to find the value of D.
With each critical point classified, the corresponding z values are f 0, 2 24 32 2 03 02 48 3 3 3 27 f 2 , 2 24 32 2 2 3 2 2 1300 48.15 When a business produces several products, a multivariable profit function may be calculated to find the production levels that maximize profit. Example 7 Maximize ProfitA small startup company produces speakers and subwoofers for computers that they sell through a website. After extensive research, the company has developed a revenue function, R x, y x 110 4.5x y 155 2 y thousand dollars where x is the number of subwoofers produced and sold in thousands and y is the number of speakers produced and sold in thousands. The corresponding cost function is C x, y 3x2 3y2 5xy 5 y 50 thousand dollars Find the production levels that maximize revenue. Solution By subtracting the cost from the revenue, we get the profit function, P x, y R x, y C x, y x 110 4.5x y 155 2 y 3x2 3y2 5xy 5 y 50 110x 4.5x2 155 y 2 y2 3x2 3y2 5xy 5 y 50 7.5x2 5 y2 5xy 110x 150 y 50 The first partial derivatives are Px x, y 15x 5 y 110 Py x, y 10 y 5x 150 The critical point are found by setting the partial derivatives equal to zero. This results in a system of equations that is solved using the Elimination Method. 15x 5 y 110 0 15x 5 y 110 10 y 5x 150 0 5x 10 y 150 Multiply the second equation by -3 and add it to the first equation to eliminate x: 15x 5 y 110 15x 30 y 450 25 y 340 y 13.6 If we substitute this value in the second equation, we get a value for x, 5x 10 13.6 150 5x 136 150 5x 14 x 2.8 The critical point is at 2.8,13.6 . This point could be a relative maximum, a relative minimum, or a saddle point. The Test for Relative Extrema helps us to distinguish whether the point is a relative maximum. The second partial derivatives are Pxx x, y 15 Pyy x, y 10 Pxy x, y 5 Each of these derivatives is a constant so any critical points have D 1510 52 125 Pxx Pyy Pxy Since D 0 and Pxx 0 , the critical point is a relative maximum. The profit at these production levels is P 2.8,13.6 7.52.82 513.62 52.813.6 110 2.8 150 13.6 50 6326 At a production level of 2.8 thousand subwoofers and 13.6 thousand speakers, the company will lose 6326 thousand dollars. The company will need to reassess their business model in order to make positive profit. Download 293.51 Kb. Do'stlaringiz bilan baham: |
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