Attatched a pic of the question much appreciated solved this maximum of 5 at (1,3)

There is a maximum value of 5 located at (x,y) = (1, 3).

Step-by-step explanation:

f(x,y) = 15 − x² − y², x + 3y = 10

Solve for x in the constraint and substitute:

x = 10 − 3y

f(y) = 15 − (10 − 3y)² − y²

Take derivative:

df/dy = -2 (10 − 3y) (-3) − 2y

df/dy = 6 (10 − 3y) − 2y

df/dy = 60 − 18y − 2y

df/dy = 60 − 20y

Set to 0 and solve:

0 = 60 − 20y

y = 3

Solve for x:

x = 10 − 3y

x = 1

Evaluate f(x,y) at (1, 3):

f(1,3) = 15 − x² − y²

f(1,3) = 15 − 1 − 9

f(1,3) = 5

Using second derivative to classify the extremum:

d²f/dy² = -20 < 0

The extremum is a maximum.

Check by comparing to another point on f(y):

f(y) = 15 − (10 − 3y)² − y²

f(0) = 15 − (10 − 0)² − 0²

f(0) = 15 − 100

f(0) = -85

This is less than the value of the function at the extremum, which confirms that it is a maximum.