1. A cylindrical can is to have a volume of 400 ... - BestWeb

1 1. A cylindrical can is to have a volume of 400 cm3. Find the dimensions (height andradius) of the can so as to minimize its total surface area . (The surface area comprisesthe top and bottom and the lateral surface.)Solution: Letrandhdenote the radius and height of the can. Here is a sketch of the canand the material used to construct rrrFrom the given information, we have400 = r2hso thath=400 r2. The lateral surface of the can is a rectangle with heighthand width 2 top and bottom of the can are disks with radiusr. The total surface areaAof the canis given byA= 2( r2) + 2 rh= 2 r2+ 2 r 400 r2= 2 r2+ need to minimizeA, given thatr >0. We haveA = 4 r 800r2 The critical numbers withr >0 will satisfy the equation4 r= only solution to this equation isr=3 200.

2 We check that whenris near zero,A (r) isnegative, and whenris large,A (r) is positive, so we know thatAhas a local and absoluteminimum atr=3 200 . The dimensions of the can with the minimum surface area arer=3 200 h=400 (200 )23= 23 200 .2. One corner of a rectangle is at the origin in thexy-plane, and the opposite corner liesin the first quadrant, along the liney= 7 23x. The sides of the rectangle are parallelto the coordinate axes. Find the largest possible area of such a :Let (x, y) be the coordinates of the rect-angle s upper right corner. Then thearea of the rectangle is given byA=xy.(x,y)We also know thaty= 7 23x, so we can takeA=x(7 23x)= 7x getA = 7 43xso thatAhas only one critical number atx= is easy to see thatA is positive for values ofxless than214and negative for values ofxgreater than214, so that we have a local maximum atx=214.

3 The area of the largestrectangle is214(7 23214)= Find the positive numberxfor which 5x+1x2is as small as : Letf(x) = 5x+1x2. We need to minimizef(x) withx >0. We getf (x) = 5 only cricital number is the solution to5 =2x3x3= is easy to check thatf (x) is negative forx <3 25andf (x) is positive forx >3 25, so wehave a local (and absolute) minimum. The functionfis minimized whenx=3 Find the points on the curvey=x2 1 that are closest to the :Let (x, y) be a point on the curve. We want tominimize the distance from (x, y) to the may as well minimize the square of the dis-tance to the origin, that isD=x2+y2.(x,y)We know thaty=x2 1, so we can writeD=x2+ (x2 1) a derivative, we getD = 2x+ 4x(x2 1)= 2x(1 + 2(x2 1))= 2x(2x2 1).

4 The critical numbers arex= 0 andx= 12. Here is a sign chart forD :- 12 1202x:2x2 1:D (x): +++ + + +There is a local maximum atx= 0 and there are local (and absolute) minima atx= points on the curve that are closest to the origin are( 12, 12)5. A rectangular garden with an area of 1000 square meters is to be laid out beside astraight river. The bank of the river provides one side of the garden; along the otherthree sides we need to put up fencing. Find the minimum amount of fencing neededto enclose the :Letxdenote the length of the side of the garden parallel tothe river and letydenote the length of the adjacent the amount of fencing we will need.

5 Then wehaveP=x+ 2y,and our goal is to minimizeP. We also know that the areaof the garden,xy, is equal to 1000, so we havexy= 1000y= foryin the expression forP, we getP=x+ find thatP = 1 the critical number ofP(withx >0) satisfies1 = 2000 = 20 5 is a critical number forP. Here is a sign chart forP :20 5 +We have an absolute minimum atx= 20 5. Thus for the minimum amount of fencing, wetakex= 20 5 andy=1000x=50 5= 10 minimum amount of fencing is given byx+ 2y= 40 5 meters.