Answer:
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.
