A rectangular storage container with an open top is to be constructed. The lengt

Question

A rectangular storage container with an open top is to be constructed. The length of the base is twice the width. Suppose that the total surface area is 600 cm^2, find the dimensions of the box that maximize the volume. A farmer wants to fence an area of 98 square feet in a rectangular field and then divide it into three cells with fences parallel to one of the sides of the rectangle. What should the lengths of the sides of the rectangular field be so as to minimize the cost of the fence or to minimize the amount of fence required

Explanation / Answer

a) let x be length of base , y be width of base , z be the height of the box

length is twice the width ==>x=2y

surface area of open top box =x*y +2zx+ 2zy =600

==>2y*y +2z*2y +2zy =600

==>2y2+6zy =600

==>z=(600-2y2)/(6y)

volume of the box v=x*y*z

v=2y*y*(600-2y2)/(6y)

v=(600y-2y3)/3

maximum volume ==> dv/dy =0 , d2v/dy2<0

dv/dy =(600-6y2)/3 =0

==>y2=600/6

==>y=10

d2v/dy2=(0-12y)/3

at y=10 ==> d2v/dy2=(0-12*10)/3 =-40<0

so maximum volume when y= 10cm

x =2y=2*10=20cm

z=(600-2y2)/(6y)=(600-2*102)/(6*10)=400/60=20/3 cm

dimenions of the box are length=20cm, width =10cm , height =20/3 cm

B) let the length of partision = length of field= x , width of field =y

area of field =x*y =98 ==> y=98/x

amount of fence required =s = (x+x+x+x)+2y

s=4x+2y

s=4x+2(98/x)

for minimum amount of fence ds/dx =0, d2s/dx2>0

ds/dx =4 -196/x2=0

==>x2=49

==>x=7

d2s/dx2=0+392/x3 for x=7 ,d2s/dx2>0

y =98/x=98/7 =14

length ofside parallel to partisian =7 ft , width of field =14 feet for minimum amount of fence

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