Question

Use the identity (x+y)(x2−xy+y2)=x3+y3 to find the sum of two numbers if the product of the numbers is 28, the sum of the squares is 65, and the sum of the cubes of the numbers is 407.

Answer 1

Answer:

Step-by-step explanation:

1) f(x) = x+y , xy = 49 --> f(x) = x+49/x

f '(x) = 1-49/x2 , x = ±√49 , choose x = +7

y = 49/x = 7

smaller = 7

larger = 7

2) Vertical Distance is difference between the two functions

V(x) = x+56 - x2, V '(x) = 1 -2x --> x = 1/2 --> V(1/2) = 56.25 <---Max Vertical distance

Boundary at -7 and 8 are less

3) I googled the equation, it is really y(n) = kn/(49+n2)

y '(n) = (k(49+n2) - 2kn2)/(49+n2) , Setting to zero, k(49+n2) -2kn2 = 0 , n = 7

4) Let x and y be the sides of the rectangle. A = xy = 37.5 Million or y = 37.5/x

The length of the fences is f(x) = 2x + 3y = 2x+112.5/x

f '(x) = 2-112.5/x2 --> x = √56.5 = 7.5 and y = 37.5/7.5 = 5 then scale by √Million

smaller = 5000 ft and larger = 7500 ft

this took me ages to work out so hopefully it helped

Answer 2

Answer:

x + y = 11

Step-by-step explanation:

xy = 28

x² + y² = 65

x³ + y³ = 407

(x + y)(x² - xy + y²) = x³ + y³

(x + y)(x² + y² - xy) = x³ + y³

(x + y)(65 - 28) = 407

37(x + y) = 407

x + y = 407÷37

x + y = 11