Answer
$54\ unit^2$
Work Step by Step
Step 1. Illustrate the configuration as shown in the figure, where $(x,y)$ is the top right vertex of the triangle satisfying $y=27-x^2, x\gt0, y\gt0$.
Step 2. The area of the isosceles triangle is given by $A=\frac{1}{2}(2x)(y)=xy=x(27-x^2)=27x-x^3$
Step 3. Take the derivative to get $A'=27-3x^2$. Letting $A'=0$, we have $x=3$ (positive answer only) and $A(3)=54\ unit^2$
Step 4. Checking $A''(3)=-6x=-18\lt0$, we can confirm the above value is a maximum.
