Question

Y = ln(17x) - 17x + 17 найти наибольшее значение функции

Answer 1

Y = ln(17x) - 17x + 17   y=1/17x-17=0    1/17x=17   17x=1/17    x=1/289
x=1/17²    ymax=-ln17-1/17+17≈17-0.059-2.83=14.11

Answer 2

To find the maximum value of the function $Y = \ln(17x) - 17x + 17$, we need to find the critical points and determine whether they correspond to a maximum or minimum.

Step 1: Find the derivative of $Y$ with respect to $x$: $Y'(x) = \frac{d}{dx}[\ln(17x) - 17x + 17]$

Using the rules of differentiation, the derivative of $\ln(17x)$ is $\frac{1}{17x}$, and the derivative of $-17x$ is $-17$: $Y'(x) = \frac{1}{17x} - 17$

Step 2: Find the critical points by setting the derivative equal to zero and solving for $x$: $\frac{1}{17x} - 17 = 0$

Solve for $x$: $\frac{1}{17x} = 17$ $x = \frac{1}{17 \cdot 17}$ $x = \frac{1}{289}$

Step 3: Determine the nature of the critical point. To check if this critical point corresponds to a maximum or minimum, we need to find the second derivative and evaluate it at $x = \frac{1}{289}$.

Step 4: Find the second derivative of $Y$ with respect to $x$: $Y''(x) = \frac{d^2}{dx^2}[\frac{1}{17x} - 17]$

Using the rules of differentiation, the derivative of $\frac{1}{17x}$ is $-\frac{1}{17x^2}$, and the derivative of $-17$ is $0$: $Y''(x) = -\frac{1}{17x^2}$

Step 5: Evaluate the second derivative at $x = \frac{1}{289}$: $Y''(\frac{1}{289}) = -\frac{1}{17(\frac{1}{289})^2} = -\frac{1}{17} \cdot 289^2 = -\frac{1}{17} \cdot 83521$

Since the second derivative is negative, the critical point at $x = \frac{1}{289}$ corresponds to a local maximum.

Step 6: Check for boundary points: The function $Y$ involves a natural logarithm $\ln(17x)$. To ensure the logarithm is defined, $x$ must be greater than 0. Therefore, there are no boundary points to consider in this case.

Step 7: Find the maximum value of the function $Y$: To find the maximum value of $Y$, substitute the critical point $x = \frac{1}{289}$ into the original function: $Y(\frac{1}{289}) = \ln(17 \cdot \frac{1}{289}) - 17 \cdot \frac{1}{289} + 17$

Calculate the value: $Y(\frac{1}{289}) = \ln(\frac{17}{289}) - \frac{17}{289} + 17$

Using a calculator, $\ln(\frac{17}{289}) \approx 3.6931$, and the calculation becomes: $Y(\frac{1}{289}) \approx 3.6931 - 0.0589 + 17 \approx 20.6342$