Question

You are designing the top of a pencil pouch that consists of a rectangle to which a semicircle has been attached to both ends. The semicircle will be made from material that costs 1ct per square centimeter. The material of the rectangular part costs 4ct per square centimeter. The perimeter of the pouch must be 80cm. How do you have to choose the radius and the 'open' side of the rectangle so the cost is minimal? Before you start, label your sketch! Please write down strategy and final result.

Answer 1

Solution:-

Let the diameter of the semicircle be d and radius be r.Let one side of the rectangle be a and other being d(=2r).(as shown in figure)

(a)

Now according to question, permeterpeof rectangle is 80 cm.

So, 2(a+d)=80

Or 2(a+2r)=80

Or a+2r=40

Or a=40-2r ..........1

Now,the area of the rectangle =a×d=a×2r=2ar

Since cost of rectangle per cm2 is 4ct.

So the Cost of area 2ar =4×(2ar)=8ar

Putting the value of a from equation 1, we get

the cost of rectangle =8(40-2r)r .........2

Now, the area of semicircle is (πr×r)/2

So area of 2 semi circles is πr×r.

Since the cost of 1 cm2 of semicircle is 1ct

So the cost of (πr×r) cm2 of semicircle=π×r×r ....3

So from equations 2 and 3 ,we get

Total cost=8(40-2r)r + π×r×r

Or total cost=320r -16r×r + π×r×r

Or total cost=320r-(16-π)r×r

Now to get minimum tatal minimum cost , let us differentiate it with respect to r and put it equals to 0,we get

Or

Or 320-(12.86)×2r =0

Or 320-25.72r=0

Or 25.72r=320

Or r=320/25.72

Or r=12.44cm

Open side length=a=40-2r=40-2(12.44)=40-24.88= 15.12cm

Hence, the radius of the semicircle and the open side of the rectangle are 12.44cm and 15.12cm respectively.