Maximizing revenue A sales analyst determines that the revenue from sales of fruit smoothies is giv?en? by ?? ?)= ?60?x2? + 300?x? here ?x is the price in dollars charged per item, w ? ith 0 ?? ? 5. a. Find the critical points of the revenue function. b. Determine the absolute maximum value of the revenue function and give the price that maximizes the revenue.

STEP_BY_STEP SOLUTION Step-1 Let f be a continuous function defined on an open interval containing a 1 number ‘c’.The number ‘c’ is critical value ( or critical number ). If f (c) 1 = 0 or f (c) is undefined. A critical point on that graph of f has the form (c,f(c)). Step-2 When an output value of a function is a maximum or a minimum over the entire domain of the function, the value is called the absolute maximum or the absolute minimum. Let f be a func tion with domain D and let c be a fixed constant in D . Then the output value f( ) is the 1. Absolute maximum value of f on D if and only if f(x) f(c) , for all x in D. 2. Absolute minimum value of f on D if and only if f(c) f(x) , for all x in D. Step_3 a) .Given is ; A sales analyst determines that the revenue from sales of fruits 2 smoothies is R(x) = -60x + 300x , here x is the price in dollars charged per item , with 0 x 5. Now , we have to find the critical points of the revenue function.For that we have to evaluate the function at critical points.so, the critical points satisfies the equation .Here the given function is polynomial function and it is continuous for all of x. Given function is; R(x) = -60x + 300x . For the critical points we have to differentiate the function with respect to x .[ Since from the step-1] d d 2 dx (R(x)) = dx( -60x + 300x ) = d ( -60x ) + d (300x ) ,since d (u+v) = d (u)+ d (v) dx dx dx dx dx d 2 d d = -60 dx( x ) + 300 (xdx),s ince dx (Cx) = C = -60(2x) +300(1) , since d (x ) = n x n1 dx = -120x+300. Since , from the definition f (c) = 0 = -120c+300 That is , -120c+300 = 0 -60(2c-5) =0. 2c-5 = 0 5 2c = 5 c = 2 . Therefore, c = 2 is clearly lies in the given interval [0,5]. 5 2 Therefore , at x = (5/2) then R( 2 ) = -60(5/2) + 300(5/2) = -60( 25) + 150(5) 4 = -375+750. = 375. 5 Hence, the function attains the critical value at x = 2 , and the critical point is ( , 375). Step_4 b). Now , we have to determine the absolute extreme values of f on the given interval ; 5 Here the critical value is 2 , and the given interval is [0,5] .So , the endpoints are 0,5 ince from the step -2, the absolute extreme values are ; 2 Therefore , at x = 0 then R(0) = - 60(0) + 300(0) = 0. 5 2 At x = (5/2) then R( 2 ) = -60(5/2) + 300(5/2) = -60( 25) + 150(5) 4 = -375+750. = 375. At x = 5 , then R (5) = -60(5) + 300(5) = -60(25) +1500 = -1500+1500 = 0. Therefore , R(0) = 0 , R( ) = 375 , and R(5) = 0. 2 5 Hence , the largest value of R(x) is 375 ,this attains at x = 2 Therefore , the absolute maximum is R( ) =375 2 Hence , the smallest value of R(x) is 0,this attains at x= 0 , 5 Therefore , the absolute minimum is R(0) = 0 = R(5) 5 Therefore , the maximum value occurs at x= 2 .Hence , absolute maximum value of the revenue function is 5 , and the maximize price is 375. 2