$$ y=\sqrt{20 x-x^{2}}+2 $$ where $$ 5 \leq x \leq 20 $$

Question

UpStudy AI · Accepted Answer

To analyze the function $$ y = \sqrt{20x - x^2} + 2 $$ for the interval $$ 5 \leq x \leq 20 $$, we first need to determine the domain of the square root function. The expression inside the square root, $$ 20x - x^2 $$, must be non-negative: $$ 20x - x^2 \geq 0 $$ Rearranging this gives: $$ -x^2 + 20x \geq 0 $$ Factoring out a negative sign: $$ -x(x - 20) \geq 0 $$ This inequality holds when $$ x(x - 20) \leq 0 $$. The critical points are $$ x = 0 $$ and $$ x = 20 $$. The sign of the expression changes at these points. We can analyze the intervals: 1. For $$ x < 0 $$: $$ x(x - 20) > 0 $$ 2. For $$ 0 < x < 20 $$: $$ x(x - 20) < 0 $$ 3. For $$ x > 20 $$: $$ x(x - 20) > 0 $$ Thus, the expression $$ 20x - x^2 $$ is non-negative for $$ 0 \leq x \leq 20 $$. Now, since we are interested in the interval $$ 5 \leq x \leq 20 $$, we can conclude that $$ 20x - x^2 $$ is non-negative in this range. Next, we can find the maximum value of $$ y $$ by analyzing the function $$ y = \sqrt{20x - x^2} + 2 $$. To find the maximum of $$ \sqrt{20x - x^2} $$, we can rewrite it as: $$ y - 2 = \sqrt{20x - x^2} $$ Squaring both sides gives: $$ (y - 2)^2 = 20x - x^2 $$ Rearranging this leads to: $$ x^2 - 20x + (y - 2)^2 = 0 $$ This is a quadratic equation in $$ x $$. For $$ x $$ to have real solutions, the discriminant must be non-negative: $$ D = b^2 - 4ac = (-20)^2 - 4(1)((y - 2)^2) \geq 0 $$ Calculating the discriminant: $$ 400 - 4(y - 2)^2 \geq 0 $$ This simplifies to: $$ 100 \geq (y - 2)^2 $$ Taking the square root of both sides gives: $$ 10 \geq |y - 2| $$ This leads to two inequalities: $$ -10 \leq y - 2 \leq 10 $$ Adding 2 to all parts: $$ -8 \leq y \leq 12 $$ Since $$ y = \sqrt{20x - x^2} + 2 $$ and $$ \sqrt{20x - x^2} $$ is always non-negative, the minimum value of $$ y $$ occurs when $$ x = 5 $$: $$ y(5) = \sqrt{20(5) - 5^2} + 2 = \sqrt{100 - 25} + 2 = \sqrt{75} + 2 = 5\sqrt{3} + 2 \approx 8.66 $$ The maximum value occurs when $$ x = 20 $$: $$ y(20) = \sqrt{20(20) - 20^2} + 2 = \sqrt{400 - 400} + 2 = 0 + 2 = 2 $$ Thus, the function $$ y $$ achieves its maximum value at $$ x = 10 $$: $$ y(10) = \sqrt{20(10) - 10^2} + 2 = \sqrt{200 - 100} + 2 = \sqrt{100} + 2 = 10 + 2 = 12 $$ In summary, the function $$ y = \sqrt{20x - x^2} + 2 $$ for $$ 5 \leq x \leq 20 $$ has a minimum value of $$ 2 $$ and a maximum value of $$ 12 $$. The function $$ y = \sqrt{20x - x^2} + 2 $$ for $$ 5 \leq x \leq 20 $$ has a minimum value of $$ 2 $$ and a maximum value of $$ 12 $$.

where

Upstudy AI Solution

Answer

Solution

Beyond the Answer

Related Questions

Latest Pre Calculus Questions