A companys cost function is C(x) = 16x^2 + 900 dollars, where x is the number of units. Find the marginal cost function. Marginal Cost is the derivative of the Cost function. [B]C'(x) = 32x[/B]

Free Derivatives Calculator - This lesson walks you through the derivative definition, rules, and examples including the power rule, derivative of a constant, chain rule

Determine whether the statement is true or false. If y = e^2, then y’ = 2e e^2 is a constant, and the derivative of a constant is 0. So y' = 0 So this is [B]FALSE[/B]

Free Functions-Derivatives-Integrals Calculator - Given a polynomial expression, this calculator evaluates the following items:
1) Functions ƒ(x).  Your expression will also be evaluated at a point, i.e., ƒ(1)
2) 1^st Derivative ƒ'(x)  The derivative of your expression will also be evaluated at a point, i.e., ƒ'(1)
3) 2^nd Derivative ƒ''(x)  The second derivative of your expression will be also evaluated at a point, i.e., ƒ''(1)
4)  Integrals ∫ƒ(x)  The integral of your expression will also be evaluated on an interval, i.e., [0,1]
5) Using Simpsons Rule, the calculator will estimate the value of ≈ ∫ƒ(x) over an interval, i.e., [0,1]

a) 1800 is the cost to run the business for a day. To clarify, when you plug in x = 0 for 0 candy bars sold, you are left with -1,800, which is the cost of doing business for one day. b) Maximum profit is found by taking the derivative of the profit equation and setting it equal to 0. P'(x) = -0.002x + 3 With P'(x) = 0, we get: -0.002x + 3 = 0 Using our [URL='http://www.mathcelebrity.com/1unk.php?num=-0.002x%2B3%3D0&pl=Solve']equation solver[/URL], we get: x = 1,500 To get maximum profit, we plug in x = 1,500 to our [I]original profit equation[/I] P(1,500) = ? 0.001(1,500)^2 + 3(1,500) ? 1800 P(1,500) = -2,250 + 4,500 - 1,800 P(1,500) = $[B]450[/B]

If 800 feet of fencing is available, find the maximum area that can be enclosed. Perimeter of a rectangle is: 2l + 2w = P However, we're given one side (length) is bordered by the river and the fence length is 800, so we have: So we have l + 2w = 800 Rearranging in terms of l, we have: l = 800 - 2w The Area of a rectangle is: A = lw Plug in the value for l in the perimeter into this: A = (800 - 2w)w A = 800w - 2w^2 Take the [URL='https://www.mathcelebrity.com/dfii.php?term1=800w+-+2w%5E2&fpt=0&ptarget1=0&ptarget2=0&itarget=0%2C1&starget=0%2C1&nsimp=8&pl=1st+Derivative']first derivative[/URL]: A' = 800 - 4w Now set this equal to 0 for maximum points: 4w = 800 [URL='https://www.mathcelebrity.com/1unk.php?num=4w%3D800&pl=Solve']Typing this equation into the search engine[/URL], we get: w = 200 Now plug this into our perimeter equation: l = 800 - 2(200) l = 800 - 400 l = 400 The maximum area to be enclosed is; A = lw A = 400(200) A = [B]80,000 square feet[/B]

The function f(x) = e^x(x - 3) has a critical point at x = The critical point is where the derivative equals 0. We multiply through for f(x) to get: f(x) = xe^x - 3e^x Using the product rule on the first term f'g + fg', we get: f'(x) = xe^x + e^x - 3e^x f'(x) = xe^x -2e^x f'(x) = e^x(x - 2) We want f'(x) = 0 e^x(x - 2) = 0 When [B]x = 2[/B], then f'(x) = 0

The function P(x) = -30x^2 + 360x + 785 models the profit, P(x), earned by a theatre owner on the basis of a ticket price, x. Both the profit and the ticket price are in dollars. What is the maximum profit, and how much should the tickets cost? Take the [URL='http://www.mathcelebrity.com/dfii.php?term1=-30x%5E2+%2B+360x+%2B+785&fpt=0&ptarget1=0&ptarget2=0&itarget=0%2C1&starget=0%2C1&nsimp=8&pl=1st+Derivative']derivative of the profit function[/URL]: P'(x) = -60x + 360 We find the maximum when we set the profit derivative equal to 0 -60x + 360 = 0 Subtract 360 from both sides: -60x = -360 Divide each side by -60 [B]x = 6 <-- This is the ticket price to maximize profit[/B] Substitute x = 6 into the profit equation: P(6) = -30(6)^2 + 360(6) + 785 P(6) = -1080 + 2160 + 785 [B]P(6) = 1865[/B]

The height of an object t seconds after it is dropped from a height of 300 meters is s(t)=-4.9t^2 +300. Find the average velocity of the object during the first 3 seconds? (b) Use the Mean value Theorem to verify that at some time during the first 3 seconds of the fall the instantaneous velocity equals the average velocity. Find that time. Average Velocity: [ f(3) - f(0) ] / ( 3 - 0 ) Calculate f(3): f(3) = -4.9(3^2) + 300 f(3) = -4.9(9) + 300 f(3) = -44.1 + 300 f(3) = 255.9 Calculate f(0): f(0) = -4.9(0^2) + 300 f(0) = 0 + 300 f(0) = 300 So we have average velocity: Average velocity = (255.9 - 300)/(3 - 0) Average velocity = -44.1/3 Average velocity = -[B]14.7 [/B] Velocity is the first derivative of position s(t)=-4.9t^2 +300 s'(t) = -9.8t So we set velocity equal to average velocity: -9.8t = -14.7 Divide each side by -9.8 to solve for t, we get [B]t = 1.5[/B]