Question

if x^2 + (y-3)^2=9, then cube root of the maximum value of (3x+4y) is​

Answer 1

Answer:

cube root of the maximum value of (3x+4y) is​ 3

Step-by-step explanation:

If x^2 + (y-3)^2=9, then cube root of the maximum value of (3x+4y) is​

x² + (y -3)² = 9

=> x² = 9 - (y -3)²

=> x = √(9 - (y -3)²)

3x + 4y

= 3(√(9 - (y -3)²)) + 4y

Differentatiting wrt y

= (3/2) (-2(y-3))/√(9 - (y -3)²)  + 4

= -3(y-3)/√(9 - (y -3)²)  + 4

Equating with 0

-3(y-3)/√(9 - (y -3)²) = -4

=> 3(y-3) = 4√(9 - (y -3)²)

Squaring both sides

=> 9(y-3)² = 16(9 - (y -3)²))

=> 25(y-3)² = 144

=>(y-3)² = 144/25

=> y - 3 = 12/5

=> y = 27/5

x² + (y -3)² = 9

=> x² + 144/25 = 9

=> x² = 81/25

=> x = 9/5

3x + 4y  =  3(9/5)  + 4(27/5)

= 135/5

= 27

= 3³

∛27  = 3

cube root of the maximum value of (3x+4y) is​ 3