Please, Please Help?

Question

This is a minimum/ maximum question, any help you can give me i will be grateful. Please show me the work behind your answer.


Sum of two positive numbers is 8. Find the minimum value of the sum of the square of one and the cube of the other.

Answer 1

Let x be the first number
Let 8-x be the second number.

The equation is:
f(x) = (8-x)² + x^3
f(x) = 64 - 16x + x² + x^3

Taking the derivative:
f'(x) = 3x² + 2x - 16

Set this to zero:
0 = 3x² + 2x - 16
0 = (3x + 8)(x - 2)

x = 2
x = -8/3

Because the answer said they had to be positive, then you have 2 and 6.

The minimum value is 2^3 + 6^2 = 8 + 36 = 44

Answer 2

let x be the first number. then we know that the second number is 8-x. why - lets add them up x+(8-x)=x+8-x=8.

lets call f the square of the one and the cube of the other

f=x^2+(8-x)^3

when do you get a min or a max, when the derivative is zero. lets find the derivative and set it equal to zero

df/dx=2x-3(8-x)^2=2x-3(64-16x+x^2)=2x-192+48x-3x^2=0 or
3x^2-50x+192.

One root is 6 and the other root is 64/6. lets plug in and see what we get.

when the one root is 6 we get f=36+8=44. At the other root we get 94.8 . So on this formulation the minimum is 44 when the number is 6

lets try the other branch when f=(8-x)^2+x^3
then df/dx=-2(8-x)+3x^2=-16+2x+3x^2=0
now the roots are x=2 and -16/6. lets ignore the negative and just worry about x=2, then f=(8-2)^2+2^3=6^2+8=36+8=44

so the answer is when x=6 or x=2 and the minimum when the sum is 8 and you square one nmber and cube the other is 44

Answer 3

The sum of the two numbers is eight, so one number is x and the other is 8-x with a range for x from zero to 8.
The sum of the square of one and the cube of the other is:

y(x) = x² + (8 - x)³ for x = [0,8]

Minimize this function to find your answer. Take the derivative and set it equal to zero:

dy/dx = 2x - 3*(8 - x)²

Which has only one zero between 0 and 8, at x = 6. So

y(6) = 6² + 2³ = 36 + 8 = 44

Answer 4

let a , b be the 2 positive numbers.

a + b = 8
minimize S = a^2 +b^3

a+b =8 ----> b = 8-a
substitute this into S
S = a^2 +(8-a)^3
S = -a^3+25a^2-192a+512

find the derivative of S with respect to a
S ' = -3a^2 +50a -192 =0
Discriminant = 2500 -4(3)(192) =196 = 14^2
a1 = (-50 -14)/(-6) = 32/3
a2 = (-50 +14)/(-6) = 6

Answer:
a = 6 ----> b =8 - 6 =2
Min S = a^2 +b^3 = 36 +8 =44

a =32/3 (was rejected as it does not minimize S)

Answer 5

a + b = 8
sum a^3 + b^2 = a^3 + (8--a)^2 = a^3 + a^2 -- 16a + 64
sum' = 3a^2 + 2a -- 16 = 0 gives a = 2, b = 6
minimum sum = 2^3 + 6^2 = 8 + 36 = 44

Answer 6

x+y = 8 or y = 8-x
x^3 + y^2 = x^3 +(8-x)^2 = x^3 +x^2 -16x + 64
derivative of this is 3x^2 +2x -16 which factors to
(x-2)(3x+8)

so the minimum happens when x=2 and y=6
2^3 + 6^2 = 44

Answer 7

4 and 2

Answer 8

44 because 2 cubed is 8 and 6 squared is 36
other posibilities not minimum
1 cubed and 7 squared (50)
3 cubed and 5 squared (52)
4 cubed and 4 squared (80)

Answer 9

Let a and b be the two numbers.

a + b = 8
a = 8 - b

b^3 + (8 - b)^2 = f(b)
f'(b) = 3b^2 + 2(8-b) * -1
f'(b) = 3b^2 - 16 + 2b

3b^2 + 2b - 16 = 0
3b^2 - 6b + 8b - 16 = 0
3b(b - 2) + 8(b - 2)= 0
(3b + 8)(b - 2) = 0

b = -8/3 or 2

f"(b) = 6b + 2
f"(b) is positive when there is a minimum.

b = 2
2^3 + (8-2)^2 = 8 + 36 = 44

Minimum = 44

Answer 10

there is no work behind
6 and 2 and you will get 44