Maximize Q=xy^2, where x and y are positive numbers, such that x + y^2=10

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