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Multi-variable Functions MAT 1300 3X Summer 2011 1 FUNCTIONS OF TWO VARIABLES 1 2 Functions of Two Variables Ex: Consider a manufacturing plant which produces both car tires and shoe soles. Let t denote the number of tires produced and let s denote the number of shoe soles produced. Let C be the total cost in dollars of producing both tires and soles. Hence C depends on both t and s. C = f (t, s) Suppose, due to physical restrictions, that at most 100 tires and 500 shoe soles can be produced. Furthermore, suppose that the fixed cost for the plant is $8000, the cost of producing each tire is $40, and the cost of producing each shoe sole is $10. Find a mathematical expression for C = f (t, s) and find the domain of f (t, s). Ex: Compute f (20, 400) and f (100, 500). 1 FUNCTIONS OF TWO VARIABLES 3 A function of two variables, f (x, y), is a function which assigns a real value to every ordered pair of real numbers, (x, y), which is in the domain of f . The domain of f (x, y) is the set of all ordered pairs of real numbers, (x, y), which can be the input for our function f (x, y). Ex: What is the domain of f (x, y) = Compute the value of f (2, 1). x ln(y + 1) x2 + y 2 ? 1 FUNCTIONS OF TWO VARIABLES Ex: Given f (x, y) = 1 + √ x + y + 2, find the domain of f . 4 2 FUNCTIONS OF MANY VARIABLES 2 5 Functions of Many Variables We may have functions of more than two variables. Such functions cannot be graphed. Ex: Consider making a monthly contribution, M , at the end of each month, to a GIC with an annual interest rate r, compounded monthly. The value, V , of the GIC after T years is given by 12M r 12T V = f (M, r, T ) = 1+ −1 . r 12 a) Compute the value of a GIC with monthly contributions of $100 and an annual interest rate of 4% after 30 years. b) Compute the value of a GIC with monthly contributions of $200 and an annual interest rate of 4% after 30 years. 3 PARTIAL DERIVATIVES 3 6 Partial Derivatives The partial derivative of f with respect to x at the point (x0 , y0 ) is given by ∂z f (x0 + h, y0 ) − f (x0 , y0 ) fx (x0 , y0 ) = . = lim ∂x (x0 ,y0 ) h→0 h The partial derivative of f with respect to y at the point (x0 , y0 ) is given by ∂z f (x0 , y0 + h) − f (x0 , y0 ) fy (x0 , y0 ) = = lim . ∂y (x0 ,y0 ) h→0 h Computing Partial Derivatives: 1. To find fx (x, y) take the derivative of f with respect to x treating y as a constant. 2. To find fy (x, y) take the derivative of f with respect to y treating x as a constant. 3. To find either of these partial derivatives at a given point, we simply substitute in the point after taking the partial derivative. 4. All the rules for derivation (product rule, quotient rule, chain rule) still apply. The geometric interpretation of partial derivatives is similar to that of the ordinary derivative of a function of one variable. fx represents the slope of the surface pointing in the x direction while fy represents the slope of the surface pointing in the y direction. 3 PARTIAL DERIVATIVES For the partial derivative of the following functions 1. f (x, y) = p x2 − y 2 2. f (x, y) = 4x2 y + x3 y 3 + y + 4 3. f (x, y) = e2x+3y 4. f (x, y) = exy 5. f (x, y) = x ln(xy) 7 3 PARTIAL DERIVATIVES 8 Ex: Given f (x, y) = xey + y 2 , find fx (x, y) and fy (x, y) and compute fx (2, 1) and fy (2, 1). 3 PARTIAL DERIVATIVES Ex: Given x ln(x + y) , x2 + y 3 ∂z ∂z and compute ∂x and . ∂x (1,1) (1,0) z= find ∂z ∂x and ∂z ∂y 9 4 HIGHER-ORDER PARTIAL DERIVATIVES 4 10 Higher-order Partial Derivatives Since a function of two variables has two first derivatives, it has four second derivatives. They are fxx : take partial derivative with respect to x twice. fyy : take partial derivative with respect to y twice. fxy : take partial derivative with respect to x, then y. fyx : take partial derivative with respect to y, then x. The second-order partial derivatives of z = f (x, y) are ∂z ∂2z ∂ ∂ fxx (x, y) = ∂x = ∂x [fx (x, y)] = ∂x 2 ∂x 2 ∂ z ∂ ∂z ∂ fxy (x, y) = ∂y∂x = ∂y ∂x = ∂y [fx (x, y)] h i ∂ ∂2z ∂z ∂ [fy (x, y)] fyx (x, y) = ∂x∂y = ∂x ∂y = ∂x h i 2 ∂ z ∂z ∂ ∂ fyy (x, y) = ∂y = ∂y = ∂y [fy (x, y)] . 2 ∂y FACT - for any reasonable function, fxy = fyx . Ex: Find all the second-order partial derivatives of f (x, y) = xey − ln(x2 y). 4 HIGHER-ORDER PARTIAL DERIVATIVES Ex: Find all the second-order partial derivatives of f (x, y) = 3xy 2 + 2xy + x2 . 11 5 RELATIVE EXTREMA 5 12 Relative Extrema A function, x = f (x, y) has a relative maximum at the point (x0 , y0 ) if for all points (x, y) sufficiently close to (x0 , y0 ) we have that f (x, y) ≤ f (x0 , y0 ). A function, x = f (x, y) has a relative minimum at the point (x0 , y0 ) if for all points (x, y) sufficiently close to (x0 , y0 ) we have that f (x, y) ≥ f (x0 , y0 ). For example, f (x, y) = x2 + y 2 has a relative minimum at (0, 0). 6 CRITICAL POINTS 6 13 Critical Points Fact: Let (x0 , y0 ) be a relative extremum of f (x, y). fy (x0 , y0 ) exist then fx (x0 , y0 ) = 0 and fy (x0 , y0 ) = 0. If fx (x0 , y0 ) and Any point, (x0 , y0 ) of a function f (x, y) which satisfies fx (x0 , y0 ) = 0 and fy (x0 , y0 ) = 0 is called a critical point. Critical points are possible relative extrema of a function. Ex: Find all the critical points of the following functions: 1. f (x, y) = 2x2 + y 2 + 8x − 6y + 20 2. f (x, y) = −x2 − 5y 2 + 8x − 10y − 13 6 CRITICAL POINTS 3. f (x, y) = e−(x 2 +y 2 ) 4. f (x, y) = x2 + 6xy + 10y 2 − 4y + 4 14 6 CRITICAL POINTS 5. f (x, y) = x3 + 2y 2 + 5xy − 1. 15 7 THE SECOND DERIVATIVE TEST 7 16 The Second Derivative Test The Second Derivative Test for Functions of Two Variables: Suppose that f (x, y) has continuous second-order partial derivatives at all points near the critical point (x0 , y0 ). Let D(x, y) = fxx (x, y)fyy (x, y) − [fxy (x, y)]2 . 1. If D(x, y) > 0 and fxx (x0 , y0 ) > 0 then (x0 , y0 ) is a relative minimum of f (x, y). 2. If D(x, y) > 0 and fxx (x0 , y0 ) < 0 then (x0 , y0 ) is a relative maximum of f (x, y). 3. If D(x, y) < 0 then (x0 , y0 ) is a saddle point of f (x, y). For example, (0, 0) is a saddle point of f (x, y) = x2 − y 2 . Picture: 7 THE SECOND DERIVATIVE TEST Ex: Find all relative extrema of f (x, y) = 2x2 + y 2 + 8x − 6y + 20 17 7 THE SECOND DERIVATIVE TEST Ex: Find all relative extrema of 1 f (x, y) = x2 − 2xy + y 3 − 3y. 3 18 7 THE SECOND DERIVATIVE TEST Ex: Find all relative extrema of f (x, y) = x4 + 2x2 y 2 + y 2 . 19