Title: Help with this optimization calculus problem? Post by: ysy on Aug 23, 2012 A fence is to be built to enclose a rectangular area of 200 square feet. The fence along three sides is to be made of material that costs 6 dollars per foot, and the material for the fourth side costs 16 dollars per foot. Find the dimensions of the enclosure that is most economical to construct
dimenstions? i need to know this for my exam so a step by step would be nice Title: Help with this optimization calculus problem? Post by: RJW on Aug 23, 2012 I think the *more* costly side is 16 vice 12.
So, same derivation as above but answer is W = 10*sqrt(12/11) L = 20*sqrt (11/12) Title: Help with this optimization calculus problem? Post by: rkdgh2002 on Aug 23, 2012 In order to answer optimization problems, you need to set-up first the equation for what you need to optimize. In this problem, you need to optimize cost. Thus, if you let
l = length w = width C = cost then C = 6(l + l + w) + 16w C = 12l + 22w Note that in this problem, I assumed that the fourth side, which is more expensive to fence, is a width. Now, we also know that A = lw 200 = lw because the area is 200 sq ft. This information is important, as it will allow us the following transformation, l = 200/w which ultimately leads to the substitution C = 12(200/w) + 22w C = 2400/w + 22w Note that we need the cost to be a function affected by only 1 variable, and that is the reason for the substitution. Note that if we did not substitute for length, then we would have 2 changing variables on the right side of the cost equation: the length and the width. You know that 1 changes with the other, because they have to maintain the 200 sq ft of area. Now that we have set-up the equation where we would optimize the cost, and finished with the substitution so that it is only determined by 1 variable, we now begin to differentiate. C' = -2400/w^2 + 22 {I hope that you have no problem getting the derivative} Since the optimum value occurs when the first derivative is set to 0, we then have 0 = -2400/w^2 + 22 22w^2 = 2400 w^2 = 1200/11 w = sqrt(1200/11), or w = 20/11 * sqrt(33) {this was obtained by simplifying the one above, and by rationalizing it} To determine the length, we have l = 200/w l = 200/ [20/11 * sqrt(33)] l = 110/sqrt(33) l = 10/3 * sqrt(33) Thus, the dimensions of your rectangle is 20/11 * sqrt(33) by 10/3 * sqrt(33) Hope this helps!!!!! Title: Re: Help with this optimization calculus problem? Post by: BearPro on Aug 24, 2012 very elegant solution :)
|