A little tricky part 1

Algebra Level 5

$0\leq x\leq2;0\leq y\leq\frac{1}{2}$ $L=(2x-x^2)(y-2y^2)$

Find the sum of the maximum value of $L$ and the value of $x$ and $y$ when the equality holds. Given the conditions above.

The answer is 1.375.

1 solution

P C
Jan 22, 2016

L can be written as $L=x(2-x)y(1-2y)=\frac{1}{2}[x(2-x)2y(1-2y)]$ Using AM-GM we get $x(2-x)\leq1$ $2y(1-2y)\leq\frac{1}{4}$ $\Rightarrow L\leq\frac{1}{8}$ The equality holds when $x=1$ and $y=\frac{1}{4}$ $\Rightarrow\frac{1}{8}+\frac{1}{4}+1=\frac{11}{8}=1.375$

Nice solution!

Steven Jim - 4 years ago

A little tricky part 1

The answer is 1.375.

1 solution

0 pending reports