Suppose f(x, y) = -x^2 - 11y^2 -15, where x and y must satisfy the equation x +
ID: 3141383 • Letter: S
Question
Suppose f(x, y) = -x^2 - 11y^2 -15, where x and y must satisfy the equation x + y = 5 relative extrema of f subject to the given condition on x and y, by solving the second equation for y (or x). Substitute the result in the first equation. Thus, f is expressed as a function of one variable. Now find where relative extrema for f occur. f has a relative at (Type an ordered pair. Use integers or fractions for any numbers in the expression.) Enter your answer in the answer box and then click Check Answer.Explanation / Answer
f(x,y)=-x2-11y2-15
and x+y=5
y=5-x
substituting in f(x,y)
f(x)=-x2-11(5-x)2- 15
= -x2-11(25+x2-10x)-15
= -12x2-275+110x-15
= -12x2+110x-290
f'(x)= -24x+110
now equating the second derivative to 0
-24x+110=0
24x=110
x= 55/12
This is the critical number
Let's place the critical number on the number line and pick the test numbers 3 and 5 from the respective intervals
f'(3)= 38
f'(5)= -10
so by first derivative test the function has relative maxima at 55/12
f(55/12)= -12(55*55/(12*12))+110*(55/12)-290
= - 252.08+504.16-290
= -37.91
Hence f has relative maxima at -37.91