Denise G. answered • 10/18/19
Algebra, College Algebra, Prealgebra, Precalculus, GED, ASVAB Tutor
The approach for these would be the same. Come up with the equations, substitute to make one equation and graph it to determine min and max.
The equations:
x+y=30
2y+√x=Max or min.
Easiest approach for substitution is to work with the first equation, solve for x.
y=30-x Substitute it into the second equation.
2(30-x)+√x=Max or min
Graph this equation.
It looks like the max value is when x=0 and the minimum value is when x=30. Knowing y, you can solve for y values knowing they add to 30. Note, the graph will show negative numbers, you have to exclude these given that they tell you both numbers are positive.
Max: x=0, y=30
Min: x=30, y=0
Denise G.
Yes that approach will work for x^2. The slope will be 0 at a max or min.10/18/19
Nick K.
How would you solve it if it were x^2 as opposed to √x? I’m also assuming that when you got 2(30-x)+√x=, you could’ve taken the derivative and set that equal to 0 and found the critical points? Thank you!10/18/19