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Q: Calcukus Homework Help ( Answered,   1 Comment )
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 Subject: Calcukus Homework Help Category: Reference, Education and News > Homework Help Asked by: kennwatson-ga List Price: \$50.00 Posted: 19 Nov 2005 12:56 PST Expires: 19 Dec 2005 12:56 PST Question ID: 595215
 ```These are not the actual questions on my homework i did change them up some. If someone can help me out on this that would be great! Because for some reason my brain is not working when i try to do these. 1. Use the chain rule to find the 1st derivative of f(x)= (3+x)/(2-x), all under the square root. 2. Find the second derivative of the function g(x)= (x+1) under square root. 3. Given y^3 + x^2*y^5-x^4= 27 find A) dy/dx, (the 1st derivative of y with respect to x) B) find the slope of the tangent line to the curve at point (0,3) 4. You have a piece of square cardboard that has dimensions of 6 inches by 6 inches. Square corners of dimesion (H) are cut from the sides. The corners are folded up to make a box. A) what are the demension of (H) that will yeild a box of maximum volume? B) what is the maximum volume? the Volume of the box = (A)(H), where A is the area of its base and H is the Height/ 5) The demand function for a product is given by p(x)= 1000-x, where p is the unit price when the demand is x. the cost of producinig the product is determined to be C(x)= 3000+ 20x I must do the following- A) find the revenue function B) find the profit function C) find the price to be charged in order to maximize the profit D) find the demand when the profit is maximized 6) Using analytical methods find the all the Horizontal and Vertical Asymptotes of the following function- H(x)= 3x^3-3x^2+5x-10/5x^3-5x^2+6x```
 ```Hi!! Thank for ask to Google Answers!! 1. Use the chain rule to find the 1st derivative of f(x)= (3+x)/(2-x), all under the square root. Remember the chain rule: f(x) = g(h(x)) ==> f'(x) = g'(h(x)) * h'(x) For this problem call: g(h) = sqrt(h) = h^1/2 h(x) = (3+x)/(2-x) then: f(x) = g(h(x)) then: [ remember that (j/k)' = (j'k - jk')/k^2 ] g'(h) = 1/2 * h^(-1/2) = 1/(2*sqrt(h)) h'(x) = [1*(2-x)-(3+x)*(-1)] / (2-x)^2 = = [2-x+3+x] / (2-x)^2 = = 5/(2-x)^2 finally: f'(x) = g'(h(x)) * h'(x) = = 1/(2*sqrt(h(x))) * 5/(2-x)^2 = = (5/2) / [sqrt((3+x)/(2-x)) * (2-x)^2] = (you can stop here) = (5/2) / [sqrt(3+x) * (2-x)^(2-1/2)] = = (5/2) / [sqrt(3+x) * (2-x)^(3/2)] = = (5/2) / [sqrt((3+x)*(2-x)) * (2-x)] = = (5/2) / [(2-x)*sqrt(6-x-x^2)] = = 5/(4-2x) * 1/sqrt(6-x-x^2) ---------------- 2. Find the second derivative of the function g(x)= (x+1) under square root. Here we must use the chain rule again: g(x) = h(i(x)) where: h(i) = sqrt(i) ==> h'(i) = 1/(2*sqrt(i)) i(x) = (x+1) ==> i'(x) = 1 then: g'(x) = h'(i(x)) * i'(x) = = 1/(2*sqrt(i(x))) * 1 = = 1/(2*sqrt(x+1)) = = 1/2 * (x+1)^(-1/2) This is the first derivative of g. For the second derivative, one more time we need to use the chain rule: g'(x) = j(k(x)) where: j(k) = 1/2 * k^(-1/2) ==> j'(k) = -1/4 * k^(-3/2) k(x) = (x+1) ==> k'(x) = 1 then: g''(x) = j'(k(x)) * k'(x) = = -1/4 * k(x)^(-3/2) * 1 = = -1/4*(x+1)^(-3/2) This is the second derivative of g(x). ------------------- 3. Given y^3 + x^2*y^5-x^4= 27 find A) dy/dx, (the 1st derivative of y with respect to x): For simplicity we will call y' = dy/dx We have that: y^3 + x^2*y^5-x^4 = 27 then: (y^3 + x^2*y^5-x^4)'=(27)' ==> (y^3)'+(x^2*y^5)'-(x^4)'=0 ==> ==> 3y^2*y' + (2x*y^5 + x^2*5y^4*y') - 4x^3 = 0 ==> ==> y'*(3y^2 + x^2*5y^4) + 2x*y^5 - 4x^3 = 0 ==> ==> dy/dx = y' = (4x^3 - 2x*y^5) / (3y^2 + x^2*5y^4) B) find the slope of the tangent line to the curve at point (0,3) The slope of the tangent line to the curve at point (0,3) is the evaluation of y' = (4x^3 - 2x*y^5) / (3y^2 + x^2*5y^4) for x=0 and y=3 Slope(0,3) = (4*0^3 - 2*0*3^5) / (3*3^2 + 0^2*5*3^4) = 0 For additional reference, with examples, regarding the chain rule and its applications, including implicit differentation (this problem), please visit the following pages: "Karl's Calculus Tutor - 4.4 Derivatives: Chain Rule Applications": http://www.karlscalculus.org/calc4_4.html "Implicit Differentiation": http://www.math.ucdavis.edu/~kouba/CalcOneDIRECTORY/implicitdiffdirectory/ImplicitDiff.html "Calculus I (Math 2413) - Derivatives - Implicit Differentiation": http://tutorial.math.lamar.edu/AllBrowsers/2413/ImplicitDIff.asp --------------------- 4. You have a piece of square cardboard that has dimensions of 6 inches by 6 inches. Square corners of dimesion (H) are cut from the sides. The corners are folded up to make a box. (Tip = Draw the square cardboard and the square corners) A) what are the demension of (H) that will yeild a box of maximum volume? Volume of the box = (A)*(H), where A is the area of its base and H is the Height. The size of each side of the base is (6-2H), then A = (6-2H)^2; so we have that: Volume of the box = V = (6-2H)^2 * H = = (36 - 24H + H^2)*H = = H^3 - 24H^2 + 36H Note that 0 =< H =< 6 , and obviusly at the end points of the interval the volume is zero (no heigh or no base). We need to find a valid value of H that makes V'(H) = 0 . So we need to find V' and equal it to zero: V' = 3H^2 - 48H + 36 = 0 ==> H = [48 + sqrt(48^2 - 4*3*36)]/6 = 15.21 (invalid value) or H = [48 - sqrt(48^2 - 4*3*36)]/6 = 0.7889 (valid) This value of H makes the first derivative zero, to know if it this a maximum, the second derivative of V at this value must be negative: V''(H) = 6H - 48 ==> V''(0.7889) < 0 ==> at this value of H the Volume of the box yield its maximum possible volume. B) what is the maximum volume? Just evaluate the Volume formula at H=0.7889 . V(0.7889) = (0.7889)^3 - 24*(0.7889)^2 + 36*(0.7889) = = 13.955 cubic inches For references regarding maximums and minimums see: "The First Derivative: Maxima and Minima - HMC Calculus Tutorial": http://www.math.hmc.edu/calculus/tutorials/extrema/ "Calculus - Calculating Minimum and Maximum Values Part I": http://www.1728.com/minmax.htm "Calculus - Calculating Minimum and Maximum Values - Part II": http://www.1728.com/minmax2.htm "Calculus - Calculating Minimum and Maximum Values - Part III": http://www.1728.com/minmax3.htm --------------------- 5) The demand function for a product is given by p(x)= 1000-x, where p is the unit price when the demand is x. the cost of producing the product is determined to be C(x)= 3000+ 20x . A) find the revenue function Revenue = price per unit * quantity demanded = p*x = = (1000-x) * x = = 1000x - x^2 B) find the profit function Profit = Revenue - Cost = = (1000x - x^2) - (3000+ 20x) = = 980x - x^2 - 3000 C) find the price to be charged in order to maximize the profit First find the profit's first derivative and equal it to zero: P'(x) = 980 - 2x = 0 ==> x = 490 Use the second derivative test to check if it is a maximum or a minimum: P''(x) = -2 ==> P''(490) < 0 it is a maximum. We have found the demand at which the profit if maximized, we want to know the correspondent price, to do that just plug the value of x found in the demand function: p(x) = 1000-x ==> p(490) = 510 , this is the price to be charged in order to maximize the profit. D) find the demand when the profit is maximized We have found this value at point C), it is x=490 ------------------- 6) Using analytical methods find the all the Horizontal and Vertical Asymptotes of the following function: H(x)= 3x^3-3x^2+5x-10/5x^3-5x^2+6x Before starting with this problem please visit the following page: "Rational Functions and Asymptotes" at Richland Community College: http://www.richland.edu/james/lecture/m116/polynomials/rational.html Now we must start with definitions: The line y=k is an horizontal asymptote of a function f if f(x)-->k when x tends to infinite (positive or negative). This definition leads to the following rules for rational functions: 1) If the degree (highest power) of the numerator is larger than the degree of the denominator, then there is no horizontal asymptote. 2) If the degree of the numerator is smaller than the degree of the denominator, then the horizontal asymptote is at y = 0 (the x?axis). 3) If the degree of the numerator is equal to the degree of the denominator, then then y=A/B is the horizontal asymptote, where A is the coefficient of the highest power of the numerator and B is the coefficient of the highest power of the denominator. In this case the degree of the numerator is equal to the degree of the denominator, then y = 3/5 is the Horizontal Asymptote. DEFINITION 2: The line x=c is a vertical asymptote of a function f if: 1) If is undefined at x=c and 2) f(x) tends to positive or negative infinite as x approaches c from the left (c-) or from the right (c+). For rational functions the above definition means that equations of the vertical asymptotes can be found by finding the roots of the denominator. In this case the denominator is 5x^3-5x^2+6x, so we must work in finding its roots: 5x^3-5x^2+6x = x*(5x^2-5x+6) The discriminant of 5x^2-5x+6 is: D = 25 - 4*5*6 = 25 - 120 = -95 < 0, then 5x^2-5x+6 has no roots. So the only root of the denominator is x=0, and this is the only Vertical asymptote. --------------------------------------------------------------------------- I hope that this helps you. If you find something unclear, or (why not) a calculation error, or if you need further assistance in one topic of this question, please do not hesitate to request for an answer clarification before rate this answer. I will gladly response to your requests. Best regards, livioflores-ga```
 ```A correction needs to be made in the answer to question 4. The error is in the derivative of the volume V' in the h^2 coefficient. It should be: V' = 12h^2 - 48h + 36 = 12(h-3)(h-1) = 0 The desired solution is h = 1. The volume is then: V = [(6 - 2h)^2]h = [(6 - 2x1)^2]x1 = 4^2 x 1 = 16 Inspection shows this is a greater volume than the maximum volume above.```