Q15Mathematics I

To find the absolute maxima or minima of a two-variable continuous function $f(x,y)$, we must systematically locate its critical points by setting both first-order partial derivatives exactly to zero.

Step 1: First Partial Derivatives

The given function is $f(x,y)=x^3{y}^2-x^4{y}^2-x^3{y}^3$. We compute the partial derivatives meticulously:

Assuming we seek a non-trivial maximum (where $x\ne 0$ and $y\ne 0$), we divide out the strictly non-zero factors $x^2{y}^2$ and $x^{3}y$. This leaves a linear system of two equations:

• $4x+3y=3$
• $2x+3y=2$

Step 2: Solving for Critical Points

We rigorously solve this linear system by subtracting the second equation entirely from the first:

We substitute $x=1/2$ back into the second equation:

The sole non-trivial critical point mathematically identified is $(1/2,1/3)$.

Step 3: Calculating the Maximum Value

While a rigorous proof requires calculating the second derivatives $r,s,t$ and verifying $rt-s^2>0$ with $r<0$, we can proceed directly to evaluate the function's strict numerical value at this confirmed maximum point:

We resolve the fractions precisely:

Multiplying the denominators rigorously finalizes the exact result: