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# Minimum/Maximum Problem

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The difference between two numbers is 4. The sum of their squares is a minimum. What are the numbers? so far I have a - b = 4 and a^2 + b^2 = c (c is a minimum) then I isolated for a and plugged it into the second equation a = 4 + b then (4+b)^2 + b^2 = c this becomes (4+b)(4+b) + b^2 = c which equals 16 + 8b + 2b^2 = c then I rearranged it to 2b^2 + 8b + 16 = c and then I factored out the 2 which gives us 2(b^2 + 4b + 8) = 0 I'm having trouble factoring this by using decomposition. Any help would be appreciated.

Jun 19, 2021

#1
+1

The difference between the two numbers is 4. The sum of their squares is a minimum. What are the numbers?

a - b = 4

2 - (-2) = 4

22 + (-2)2 = 8

Jun 19, 2021
#2
+1

a -  b  =  4

b =  a - 4

a^2  +  b^2

a^2  +  ( a - 4)^2

a^2  +  a^2  -  6a  +  16

2a^2 -  8a   +   16

a^2   -  4a   +  8           take  the  derivative   and set to  0

2a  -  4  =  0

2a  =   4

a  =  2

b =  a - 4   =   2  - 4   =   -2

a  = 2

b =  -2                 produce a  min   Jun 19, 2021
#3
+1
To go from a^2 - 4a + 8 to 2a - 4 = 0, Did you complete the square? When you say to set equal to zero and take the derivative, I presume that means to set the equation equal to zero and then solve for a, but I can't decompose this equation. The answer makes sense, I just need some clarification on the step above. Thanks.
Guest Jun 19, 2021
#4
+1

Another  non-Calculus  approach

a^2  -  4a  +  8  will  be  a paraola  (in the  variable a)  that turns  upward

The "a"   coordinate  of  the  vertex =   - (-4 ) / ( 2 * 1)   = 2

This value of a  will produce   a  min on the  graph

So

Since

a -  b    =  4

Then  b =    -2   CPhill  Jun 19, 2021