Given the matrices A = [1 3 0 0 1 0 0 0 1], B = [0 0 1 0 1 0 1 0 0], and C = [-1

Question

Given the matrices A = [1 3 0 0 1 0 0 0 1], B = [0 0 1 0 1 0 1 0 0], and C = [-1 0 2 0 3 -2]. Find a. -A + 2B b. CA c. A^-1 d. X, solution of AX = B Using the simple method, find the maximum value of z = x + y + 7, subject to the constraints: 2x + y lessthanorequalto 6, y - x lessthanorequalto 1, x + 2y lessthanorequalto 4, and x greaterthanorequalto 0, y greaterthanorequalto 0. Solve the equations. a. 2^-4x - 5 = 0.5 b. log_3 (4x - 3) + 1 = 0 Find the present value if $800 is deposited regularly at the end of each month at 6% for 12 years. What is the APR (annual percentage rate) here? A business model has the revenue and cost function R(x) = 50x - x^2, C(x) = 10x + 300, respectively, for producing x units of items. Find a. the maximum profit b. the break-even point c. the marginal profit at x = 20 The cost of manufacturing a molded item is related to the quantity of parts produced during a When 10 parts arc produced, the cost is $30; When 60 parts are produced, the cost is $130. Assume a model. Find a. the rate of change b. the marginal cost c. the fixed cost d. the linear model function a. Find an equation of the tangent line of f(x) = -x^3 + 7.5x^2 - 12x + 3, at x = 0. b. Determine the location of each local extremum of f(x).

Explanation / Answer

5. a>

THe profit function is given as , P(x) = R(x) - C(x) = (50x-x^2) - (10x+300)

                                                                                        = -x^2 + 40x - 300

For maximum profit we'll find P'(x) first

=> P'(x) = -2x + 40

Now we'll find the critical point by solving P'(x) = 0

=> -2x + 40 = 0 , => x = 20

P(x) = -x^2 + 40x - 300
P(x=20) = -(20)^2 + 40(20) - 300 = 100

As P(x) > 0    at x = 20

=> we'll get a maxima at x = 20

and the maximum profit is P(20) = 100

b> That value of x at which the total cost is equal to the total revenue is know as the break even point.

=> C(x) = R(x)

(10x+300) = (50x-x^2)

x^2 - 40x + 300 = 0

x^2 - 10x - 30x + 300 = 0

(x-10)(x-30) = 0

=> at x = 10 and x = 30 units we'll get the break even points

c> Marginal profit is the derivative of the profit function at some point x

here x = 20

=> From part (a) P'(x) = -2x + 40
=> P'(x=20) = -2(20) + 40 = 0

=> the marginal profit at x = 20 is 0

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