ind critical points by differentiating $p(t)$

First, differentiate $p(t)$ with respect to $t$:

Let

$$ f(t) = \frac{1000t}{100 + t^2} $$

So,

$$ p'(t) = 0 + \frac{d}{dt} \left( \frac{1000t}{100 + t^2} \right) $$

Using quotient rule:

$$ \frac{d}{dt}\left(\frac{u}{v}\right) = \frac{v \frac{du}{dt} - u \frac{dv}{dt}}{v^2} $$

where $u = 1000t$, $v = 100 + t^2$.

$$ \frac{du}{dt} = 1000, \quad \frac{dv}{dt} = 2t $$

So,

$$ p'(t) = \frac{(100 + t^2)\cdot 1000 - 1000t \cdot 2t}{(100 + t^2)^2} $$

Simplifying numerator:

$$ (100 + t^2)1000 - 2000t^2 = 1000(100 + t^2 - 2t^2) = 1000(100 - t^2) $$

So,

$$ p'(t) = \frac{1000(100 - t^2)}{(100 + t^2)^2} $$

Step 2: Set $p'(t) = 0$ and solve for $t$:

$$ 1000(100 - t^2) = 0 \implies t^2 = 100 \implies t = 10 \quad (\text{since } t \geq 0) $$

Step 3: Check endpoints and critical point

For $t \to 0$,

$$ p(0) = 1000 + \frac{0}{100} = 1000 $$

For $t = 10$,

$$ p(10) = 1000 + \frac{1000 \cdot 10}{100 + 10^2} = 1000 + \frac{10000}{200} = 1000 + 50 = 1050 $$

For $t \to \infty$,

$$ p(\infty) = 1000 + \frac{1000 \cdot \infty}{100 + \infty^2} = 1000 + \frac{\infty}{\infty} $$

The degree of numerator and denominator is the same (both degree 1 in $t$ in numerator and degree 2 in denominator), so for large $t$:

$$ \frac{1000t}{t^2} = \frac{1000}{t} \to 0 $$

So,

$$ p(\infty) \to 1000 $$

Step 4: Conclusion

Maximum value is at $t = 10$

So, the maximum size is:

$$ \boxed{1050} $$

Option C is correct.

Application Of Derivatives

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