1. Interpret the expression and the line

We need to maximize $f(x,y)=xy^4$ subject to the condition that $(x,y)$ lies on the line through $(50+\alpha,0) \quad \text{and} \quad (0,50+\alpha), \qquad \alpha>0.$

The equation of the line joining these intercepts is $x+y=50+\alpha.$

So the constraint is $y=50+\alpha-x.$

2. Reduce to a single-variable function

Substitute $y=50+\alpha-x$ into $f$: $f(x)=x(50+\alpha-x)^4.$

We now maximize this for $0\le x\le 50+\alpha$.

3. Differentiate

Let $k=50+\alpha$. Then $f(x)=x(k-x)^4.$

Differentiate using the product rule: $f'(x)=(k-x)^4+x\cdot 4(k-x)^3(-1).$

So, $f'(x)=(k-x)^3\big[(k-x)-4x\big].$

Thus, $f'(x)=(k-x)^3(k-5x).$

For critical points, $f'(x)=0 \implies x=k \quad \text{or} \quad x=\frac{k}{5}.$

At $x=k$, we get $y=0$, so $xy^4=0,$ which cannot be the maximum compared to interior positive values.

Hence the relevant critical point is $x=\frac{k}{5} = \frac{50+\alpha}{5}.$

Then $y=k-x = k-\frac{k}{5}=\frac{4k}{5} = \frac{4(50+\alpha)}{5}.$

4. Find the relation between $x$ and $y$

Now compare $x$ and $y$: $y=\frac{4(50+\alpha)}{5}, \qquad x=\frac{50+\alpha}{5}.$

Therefore, $y=4x.$

So the point of maximum also lies on the line $\boxed{y=4x}.$

5. Check options

• A: $y=4x$ — Correct
• B: $x=4y$ — Incorrect
• C: $y=4x+\alpha$ — Incorrect
• D: $x=4y-\alpha$ — Incorrect

6. Comparison with stored answer

Derived answer: A

Stored correct answer: A

They agree.