CALCULUS DERIVATIVES TO SOLVE PRACTICAL PROBLEMS HELP!?

Question

I need to use differentiation to find the maximum & minimum values in these practical situations.

1. a solid rectangular block has a square base. find the maximum volume if the sum of the height and any one side of the base is 12cm?

2.a man wishes to fence in a rectangular enclosure of a area 128m^2. one side of the enclosure is formed by part of a brick wall already in position. what is the least possible length of fencing required for the other three sides?

3.an open tank is to be constructed with a horizontal square base and four vertical sides. it is to have a capacity of 32m^3.find the least area of sheet metal of which it can be made?

Thank You.

Answer 1

let the side of the square base be x. let the height be h and the volume V.

V = x^2*h

but, x + h = 12

            h = 12 - x

so,

V(x) = x^2 (12 - x).

       = 12x^2 - x^3  

we need to find the maximum of V whose domain is [0, 12]. so, first, let's take the derivative,

V'(x) = 24x - 3x^2.

setting V'(x) = 0 to find the critical numbers,

           24x - 3x^2 = 0

           x (24 - 3x) = 0

           x = 0  or x = 8.

the critical numbers are 0 and 8. since the domain is a closed interval, we can use the closed interval method to find the absolute max.

V(0) = 0

V(12) = 0

V(8) = 256.

since V(8) is the largest of these values, it is the absolute maximum.

therefore, the largest possible volume is V(8) = 256.

2) let the length of part of the brick wall be y and the other side be x.

the area

x*y = 128

y = 128 / x ................(1)

the length of fence

L = 2x + y

L(x) = 2x + 128/x  [ from (1) ]

we need to find the minimum of L. so, we first take the derivative

L'(x) =2 - 128 / x^2

setting L'(x) = 0

2 - 128/x^2 = 0

128 / x^2 = 2

x^2 = 64

x = +- 8.

x = -8 is impossible because -8 is not in the domain of L.

so, x = 8 is the only critical number. in order to prove that the absolute minimum is L(8), notice that for all x < 8, L'(x) is negative, which means L is decreasing for all x < 8. for all x > 8 in the domain of L, L'(x) is positive, or L is increasing. therefore, L(8) is the absolute minimum.

y = 128/x

  = 128/8

  = 16.

therefore, the length of sides for which the minimum length of fencing is required is 8m and 16m.

3) let the side of the square base be x and the height be y.

x^2*y = 32

y = 32 / x^2.

the surface area

s = x^2 + xy + xy + xy + xy

   = x^2 + 4xy

since y = 32 / x^2,

s(x) = x^2 + 128 / x.

we need to find the absolute minimum of s.

s'(x) = 2x - 128 / x^2.

setting s'(x) = 0,

2x - 128/x^2 = 0

x = 64/x^2

x^3 = 64

x = 4.

notice that for all x < 4, s'(x) < 0 and for all x > 4, s'(x) > 0. this means that s is decreasing before 4 and increasing after 4. therefore, the absolute minimum occurs at 4.

y = 128 / x^2

   = 128 / 16

   = 8.

therefore, the minimum surface area occurs for base 4m and height 8m.

Answer 2

l,b,h are length breadth and height

then volume=l*b*h

given l=b[square base]   and b+h=l+h=12

so volume=b^2h  and b+h=12 => volume=b^2 (12-b)

for volume to be maximum its derivative with respect to breadth is minimum =0

=>dV/db=2b(12-b)-b^2=0

=>b^2=2b(12-b)

=>b=24-2b   =>b=8  =>l=8   =>h=4

2)take l and b again

given l*b=128

to find 2l+b=minimum

b=128/l

=>2l+128/l=minimum

=>2-128/l^2=0  [differentiate]

=>128/l^2=2

=>l=8

=>b=16

=>answer is =32

3)least area is got by open cube

given volume=32 = l^2*h

to find l^2+4lh=minimum [sum of area of base + 4sum of area of sides]

h=32/l^2

=>l^2+4*32/l=minimum

differentiate

2l-128/l^2=0

=>l=4

=>h=2

=>answer is 48

Answer 3

2. Let x be the width and y be the length of the fence.

Perimeter of the fence = x+2y (because of the wall, it is not 2x+2y).

x+2y=128

2y=128-x

y=64-x/2

Area = xy = x(64-x/2)

A= 64x-x^2 / 2

dA/dx =64 -2x/2 =0

x=64

d^2A/dx^2 =-1 < 0 , indicates the area is maximized.

x= 64 m (width)

64+2y=128

2y=64

y=32 m

The other three sides are 32 m, 32 m, and 64 m

1) Let the side of the square base be x cm and the height be y cm.

V=x^2y

x+y=12 (given)

y=12-x

Volume = x^2(12-x)

V=12x^2-x^3

dV/dx =12(2x)-3x^2=0

24x-3x^2=0

3x(8-x)=0

x=8

y=12-8=4

d^2V/dx^2 = 24-6x

when x=8, d^2V/dx^2 < 0, indicating V is maximized.

Maximum volume = x^2y = (8)^2 (4) = 64(4) = 256 cubic cm

Answer 4

1. x+y=12

2.  xy=128m^2

3. xyz=32m^3