X+y^2=10
minus x from both sides
y^2=10-x
sub (10-x) for y^2 in other equation
Q=x(10-x)
Q=10x-x^2
now find the maximum value
take the derititive
dQ/dx=10-2x
it is zero at x=5
below that, it is positive
after 5, it is negative
max at x=5
solve for y
y^2=10-x
y^2=10-5
y^2=5
sqrt both sides
y=√5
x=5
y=√5
the max value is 25
Thirty Five all you need to do is add 20 to 15
