Math / Calculus

Question2. (a) Find the $x$ -coordinate of the stationary point on the curve with the equation $y=18 x-4 \sqrt{x^{3}}$ (b) Hence, determine the greatest and least values of $y$ in the interval

Studdy Solution

STEP 1

1. A stationary point on a curve occurs where the derivative of the function is zero.
2. The function given is $y = 18x - 4\sqrt{x^3}$ .
3. We will find the derivative of $y$ with respect to $x$ , set it to zero, and solve for $x$ .
4. We will evaluate the function at the stationary point and at the endpoints of the given interval to find the greatest and least values of $y$ .

STEP 2

1. Differentiate the function $y = 18x - 4\sqrt{x^3}$ .
2. Find the $x$ -coordinate of the stationary point by setting the derivative to zero.
3. Evaluate the function at the stationary point.
4. Determine the greatest and least values of $y$ in the interval by evaluating $y$ at the stationary point and the endpoints of the interval.

STEP 3

Differentiate the function $y = 18x - 4\sqrt{x^3}$ .
First, rewrite the function for easier differentiation: $y = 18x - 4(x^{3/2})$
Differentiate with respect to $x$ : $\frac{dy}{dx} = 18 - 4 \cdot \frac{3}{2}x^{1/2}$
Simplify: $\frac{dy}{dx} = 18 - 6x^{1/2}$

STEP 4

Set the derivative to zero to find the stationary point: $18 - 6x^{1/2} = 0$
Solve for $x$ : $6x^{1/2} = 18$ $x^{1/2} = 3$ $x = 9$
The $x$ -coordinate of the stationary point is $x = 9$ .

STEP 5

Evaluate the function at the stationary point $x = 9$ : $y = 18(9) - 4\sqrt{9^3}$
Calculate: $y = 162 - 4\sqrt{729}$ $y = 162 - 4 \times 27$ $y = 162 - 108$ $y = 54$

STEP 6

To determine the greatest and least values of $y$ in the interval, we need the interval endpoints. However, the interval is not provided in the problem statement. Assuming we have an interval $[a, b]$ , evaluate $y$ at $x = a$ , $x = b$ , and $x = 9$ .
Since the interval is not specified, we cannot complete this step without additional information.
The $x$ -coordinate of the stationary point is $x = 9$ , and the value of $y$ at this point is $y = 54$ . The greatest and least values of $y$ in the interval cannot be determined without the interval endpoints.

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