Question

1- a child's rectangulat play yard is to be built next to the house. to make the three sides of the playpen, 24 feet of fencing are available. what should be the dimentiins of the sides to make a maximum area? Find maximum area.

2- A metal cube dissolves in acid such that an side of the cube decreases by 0.62 mm per min. How fast is the volume of the cube changing when the edge is 3 mm? volume of cube = (side)^3

3- A spherical balloon is inflated with helium at a rate of 131 pi ft^3/min. How fast is the balloon's radius increasing when the radius is 4. volume of sphere 4piR^3/3

Answer 1

1. Let x and y be length and breadth of the playground now fencing available is 24 so permitere will be 24 for three sides. As one side is house wall.

Perimeter = 2x+y =24

Gives y= 24 -2x

Now area of playground can be given as.

A= xy = x( 24-2x) = 2( 12x -x^2)

We have to maximise area so we put A' =0. Implies

A' = 2( 12- 2x) = 0 gives, x= 6

And A'' =-4 <0 so area maximise at this point.

X=6 , y = 24 - 2×6 = 12.

So maximum area is 12×6 =72.

2. Let the side of the cube be r so here given,

dr/dt= 0.62 mm/min

V= r^3

dv/dt= 3r^2 dr/dt = 3× 3× 3× 0.62 = 16.74 mm^3/ min when r=3

3.

Given dv/ dt= 131π ft^3/ min

V= ( 4/3 ) π r^3 implies dv/dt = 4π r^2 dr/dt

At r =4 , 131π = 4π ×4×4× dr/dt

Which gives dr/ dt= 131/ 64 ft/ min

dr/dt = 2.047 ft/ min (approx)

So radius is increasing with 2.047 ft/min.