wind resistance, measured in newtons, is modeled by f(x) = x³/3405000 - x²/1000 + 145097x/141875 + 1054…

wind resistance, measured in newtons, is modeled by f(x) = x³/3405000 - x²/1000 + 145097x/141875 + 1054, where 500 ≤ x ≤ 3100 is the wind velocity in meters per second. what is the minimum wind resistance on the interval? at what velocity will the wind resistance reach its minimum on this same interval? round your final answer to the nearest hundredth. answer minimum wind resistance f(x) = newtons is obtained at x = meters per second
Answer
Explanation:
Step1: Find the derivative of $f(x)$
We have $f(x)=\frac{x^{3}}{3405000}-\frac{x^{2}}{1000}+\frac{145097x}{141875}+ 1054$. Using the power - rule $\frac{d}{dx}(x^{n})=nx^{n - 1}$, we get $f^\prime(x)=\frac{3x^{2}}{3405000}-\frac{2x}{1000}+\frac{145097}{141875}$. Simplify $f^\prime(x)$ to $f^\prime(x)=\frac{x^{2}}{1135000}-\frac{x}{500}+\frac{145097}{141875}$.
Step2: Set $f^\prime(x) = 0$
To find the critical points, we set $\frac{x^{2}}{1135000}-\frac{x}{500}+\frac{145097}{141875}=0$. Multiply through by 1135000 to clear the fractions: $x^{2}-2270x + 932776=0$.
Step3: Solve the quadratic equation
Using the quadratic formula $x=\frac{-b\pm\sqrt{b^{2}-4ac}}{2a}$ for the quadratic equation $ax^{2}+bx + c = 0$. Here, $a = 1$, $b=-2270$, and $c = 932776$. Then $x=\frac{2270\pm\sqrt{(-2270)^{2}-4\times1\times932776}}{2\times1}=\frac{2270\pm\sqrt{5152900 - 3731104}}{2}=\frac{2270\pm\sqrt{1421796}}{2}=\frac{2270\pm1192.47}{2}$. We get two solutions: $x_1=\frac{2270 + 1192.47}{2}=1731.24$ and $x_2=\frac{2270-1192.47}{2}=538.77$. Both values are in the interval $[500,3100]$.
Step4: Evaluate $f(x)$ at critical points and endpoints
Evaluate $f(x)$ at $x = 500$, $x=538.77$, $x = 1731.24$ and $x = 3100$. $f(500)=\frac{500^{3}}{3405000}-\frac{500^{2}}{1000}+\frac{145097\times500}{141875}+1054$ $=\frac{125000000}{3405000}-250+\frac{72548500}{141875}+1054$ $\approx36.71-250 + 511.36+1054=1352.07$. $f(538.77)=\frac{(538.77)^{3}}{3405000}-\frac{(538.77)^{2}}{1000}+\frac{145097\times538.77}{141875}+1054$. $f(1731.24)=\frac{(1731.24)^{3}}{3405000}-\frac{(1731.24)^{2}}{1000}+\frac{145097\times1731.24}{141875}+1054$. $f(3100)=\frac{3100^{3}}{3405000}-\frac{3100^{2}}{1000}+\frac{145097\times3100}{141875}+1054$. After calculating, we find that the minimum value occurs at $x\approx538.77$ and $f(x)\approx1349.32$.
Answer:
Minimum wind resistance $f(x)=1349.32$ Newtons is obtained at $x = 538.77$ meters per second