Question

if a^2 +b^2+c^2+d^2=800 then find the maximum value of 3a+4b+5c+6d where a,b,c and d are real numbers

Answer 1

Step-by-step explanation:

3a + 4b + 5c = 2013√(2) , then, the minimum value of a^2 + b^2 + c^2 is

Answer 2

Step-by-step explanation:

3 a + 4 b +5 c+6 d

(3^2+4^2+5^2+6^2)(a^2+b^2+c^2+d^2) >=(3a+4b+5c+6d)^2

(9+16+25+36)(a^2+b^2+c^2+d^2)>=(3a+4b+5c+6d)^2

(86)(800)>= (3a+4b+5c+6d)^2

40√43