Optimization problem? Please help!!!

Question

Given two numbers with a difference of 14, find the minimum for the sum of 4 times the larger number and the square of the smaller number. 

Answer 1

Let x be the larger number

y is smaller number

 

x - y = 14

so

y = x - 14

 

The sum of 4 times larger and

the square of the smaller is:

 

4x + (x-14)²

4x + x² - 28x + 196

x² - 24x + 196

 

This graphs as a parabola opening

upward.  From quadratic equation,

The vertex is (-b/2a, f(-b/2a)).

The minimum value, m, is the 

output at the vertex ... f(-b/2a)

 

a = 1, b = -24, c = 196

 

-b/2a = 24/2 = 12

f(12) = 12² - 24(12) + 196

        = 144 - 288 + 196

        = 52

The minimum value is 52

 

 

Answer 2

The first number we'll make "x", the second number we'll make "x+14" which is the larger of the two numbers.

 

Find the minimum for the sum of 4 times the larger number and the square of the smaller number. 

 

This statement gives this equation:

 

f(x) = 4(x+14) + x²

 

The equation is that of a parabola, so the minimum will be the vertex. For a parabola in the form y = ax² + bx +c, the vertex is located at coordinates (-b/2a , f(-b/2a)).

 

f(x) = x² + 4x + 56

 

a = 1

b = 4

c = 56

-b/2a = -4/2(1) = -2

f(-2) = (-2)² + 4(-2) + 56

f(-2) = 4 - 8 + 56

f(-2) = 52

 

vertex: (-2, 52)

 

Minimum sum of the two numbers is 52