Differentiation

Edexcel IAL WMA12/01 /P2/June 2021/Q9 (Differentiation Application- Optimization)

Figure 3 shows a sketch of a square based, open top box.​​ 

The height of the box is h cm, and the base edges each have length l cm.​​ 

Given that the volume of the box is 250000cm3​​ 

• show that the external surface area, Scm2, of the box is given by​​ 

S=250000h+ 2000 h

(3)​​ 

• Use algebraic differentiation to show that S has a stationary point when​​ h = 250 k​​ where k is a rational constant to be found.​​ 

(5)​​ 

• Justify by further differentiation that this value of h gives the minimum external surface area of the box.​​ 

(2)

SOLUTION

a-​​ Use the formula of surface area to show the given expression.​​ 

(Total Surface Area is the sum of the areas of all the faces of the prism. In this case, cuboid have 5 faces. The area of all (4) side faces is​​ 2lh​​ and bottom face is​​ l2).​​ 

S=4lh+l2

S=l2+4hl

​​ (The given surface area expression that needs to be shown has no ‘length’​​ parameter in it, so use the given information of volume to substitute the value of l in terms of h in the above expression to show the asked expression.) ​​ 

V=w x l x h

V=l×l×h

V=l2h

250000=l2h

l=250, 000h

​​ (Now, substituting the value of l in terms of h in the expression derived of surface area.)

S=l2+4hl

S =250, 000h2+4h 250, 000h

S= 250, 000h+2000 h  

b-​​ 

At stationary points, the gradient of the curve is equal to zero. Therefore, we will diffrentiate the equation of curve and equate the diffrentiated expression to zero.

S=250,000 h-1+2000 h12

dsdh= -250, 000 h-2+1000 h-12

Now,​​ 

dsdh= 0

- 250, 000 h-2+1000 h-12 =0 

-250000h2+1000h12=0

250000h2=1000h12

250h2=1h12 
250=h2h12

250=h2-12

250=h32

h32=250

h3223=25023

h=25023

-250 h2h2+h2h12= 0 x h2

-250+h32=0

h3223=25023

h=25023 

Hence, on comparing it with​​ h = 250 k​​ ,​​ k=23.​​ 

c-​​ 

(If a function​​ f(x)​​ has a stationary point when​​ x=a, then if​​ f’’a>0, the point is a local minimum.)​​ 

dsdh= -250, 000 h-2+1000 h-12

Diffrentiating again.​​ 

d2Sdh2=500, 000 h-3-500 h-3

d2Sdh2=500, 000h3-500h32

Substituting​​ h=25023

d2Sdh2=500,000250233=5002502332

d2Sdh2=500, 0002502-500250

   d2sdh2=6>0

Thus,​​ the stationary point at​​ h=25023​​ is local minimum.​​