Parabola word problem

Question

Find two or more numbers having a sum of 50 such that the sum of the square of one and three times the square of the other is a minimum.

Answer 1

X + Y = 50

 

X^2 + 3Y^2 for a minimum

 

(50-Y)^2 + 3y^2

 

2500 -100Y + Y^2 + 3y^2

 

4Y^2 -100Y + 2500 for a minimum

 

take the derivative

 

8Y -100

 

for a zero, y = 12.5

 

check the original by substituting 12.5 for Y and (50-12.5 =) 37.5 for X. 

 

answer,  1875.

 

substituting 12/38 and 13/37 result in 1876. check.  

 

 

Answer 2

x and y are the numbers

x+y=50

x=50-y

x²+3y²=minimum

substitute

(50-y)²+3y²=0

2500-100y+y²+3y²=0

2500-100y+4y²=0

divide by 4

625-25y+y²=0

y²-25y=-625

(-25/2)²=625/4

y²-25y+(625/4)=-625+(625/4)

[y-(25/2)]²=-468.75

[y-(25/2]²+468.75=0

this is a parabola opening upwards so you have a minimum value of k=468.75 when h=25/2

25/2=12.5

50-12.5=37.5

the two numbers are 12.5 and 37.5

37.5²+3(12.5)²=1406.25+468.75=1875

try 12.4 and 37.6:

37.6²+3(12.4)²=1413.76+461.28=1875.04

try 12.6 and 37.4:

37.4²+3(12.6)²=1398.76+476.28=1875.04

try 12 and 38: you get 1876

try 13 and 37: you get 1876

try 11 and 39: you get 1884

try 14 and 36: you get 1884