My Math Forum 2 questions in my math analysis hmw.

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 October 26th, 2013, 04:54 PM #1 Member   Joined: Feb 2013 Posts: 44 Thanks: 0 2 questions in my math analysis hmw. The management of the UNICO department store has decided to enclose a 906 ft² area outside the building for displaying potted plants and flowers. One side will be formed by the external wall of the store, two sides will be constructed of pine boards, and the fourth side will be made of galvanized steel fencing. If the pine board fencing costs $7/running foot and the steel fencing costs$2/running foot, determine the dimensions of the enclosure that can be erected at minimum cost. (Round your answers to one decimal place.) So we use the equations they gave us. A = xy P = 2x + y I already set 906 = xy and solved for y. then plugged it into for P and solved for x. and that's where I got lost. 2, The demand for motorcycle tires imported by Dixie Import-Export is 49,000/year and may be assumed to be uniform throughout the year. The cost of ordering a shipment of tires is $360, and the cost of storing each tire for a year is$2. Determine how many tires should be in each shipment if the ordering and storage costs are to be minimized. (Assume that each shipment arrives just as the previous one has been sold.) No idea what to do for this one.
 October 26th, 2013, 07:06 PM #2 Global Moderator   Joined: Dec 2006 Posts: 20,101 Thanks: 1905 The fencing (in \$) costs 14x + 2y = 14x + 2A/x = (?(14x) - ?(2A/x))² + 2?(14)?(2A), which is minimized when ?(14x) = ?(2A/x) (which requires x = ?(A/7), so y = A/x = ?(7A)). The question gives the value of A.
 October 26th, 2013, 08:08 PM #3 Member   Joined: Feb 2013 Posts: 44 Thanks: 0 Re: 2 questions in my math analysis hmw. So what's the answer? None of that made any sense to me. lol
 October 26th, 2013, 10:43 PM #4 Senior Member     Joined: Jul 2010 From: St. Augustine, FL., U.S.A.'s oldest city Posts: 12,209 Thanks: 517 Math Focus: Calculus/ODEs Re: 2 questions in my math analysis hmw. 1.) [color=#BF0000]skipjack[/color] used a technique that does not require the calculus. If you let $0 be the length of the two sides perpendicular to the wall and $0 be the length of the side parallel to the wall (both in feet), then your cost function in dollars is: $C(x,y)=7(2x)+2(y)=14x+2y$ We are constrained by the fact that the rectangular area is: $A(x,y)=xy=906$ Hence: $y=\frac{906}{x}$ And so we may express the cost function in terms of one variable x: $C(x)=14x+1812x^{-1}$ What you want to do now is equate the derivative to zero to find the critical value for x. You are probably expected to use either the first or second derivative tests to demonstrate that this critical value is a relative maximum.
 October 27th, 2013, 04:37 AM #5 Global Moderator   Joined: Dec 2006 Posts: 20,101 Thanks: 1905 Using calculus, one first needs to find the critical point (there happens to be only one for x > 0), and then show that its a relative minimum. It's then easy to see that it's the absolute minimum as well.

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