Write the criteria for the function y = f(x) to have the local maxima and local minima at a point. Find the local maxima and local minima of f(x) = 2x^3 15x^2 + 36x + 10 . Also find the point of…

Model Answer: NEB +2 Science Mathematics (2076) – 6 Marks

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Part 1: Local Maxima, Local Minima, and Point of Inflection

Criteria for Local Maxima and Local Minima

For a function $y=f(x)$ to have local maxima or local minima at a point $x=a$, the following conditions must be satisfied:

1. First Derivative Test:

   • If $f^{\prime }(a)=0$ or $f^{\prime }(a)$ is undefined, then $x=a$ is a critical point.
   • To determine whether it is a local maximum, local minimum, or neither:
     • If $f^{\prime }(x)$ changes from positive to negative as $x$ passes through $a$, then $x=a$ is a local maximum.
     • If $f^{\prime }(x)$ changes from negative to positive as $x$ passes through $a$, then $x=a$ is a local minimum.
     • If $f^{\prime }(x)$ does not change sign, then $x=a$ is neither a local maximum nor a local minimum.

2. Second Derivative Test (if applicable):

   • If $f^{\prime }(a)=0$ and $f^{\prime \prime }(a)<0$, then $x=a$ is a local maximum.
   • If $f^{\prime }(a)=0$ and $f^{\prime \prime }(a)>0$, then $x=a$ is a local minimum.
   • If $f^{\prime \prime }(a)=0$, the test is inconclusive.

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Finding Local Maxima and Minima for $f(x)=2x^3-15x^2+36x+10$

Step 1: Find the first derivative $f^{\prime }(x)$ $$f(x)=2x^3-15x^2+36x+10$$ $$f^{\prime }(x)=\frac{d}{dx}(2x^3-15x^2+36x+10)=6x^2-30x+36$$

Step 2: Find critical points by solving $f^{\prime }(x)=0$ $$6x^2-30x+36=0$$ Divide by 6: $$x^2-5x+6=0$$ Factorize: $$(x-2)(x-3)=0$$ Thus, critical points are at: $$x=2\text{and}x=3$$

Step 3: Determine nature of critical points using the second derivative test Find the second derivative: $$f^{\prime \prime }(x)=\frac{d}{dx}(6x^2-30x+36)=12x-30$$

Evaluate $f^{\prime \prime }(x)$ at critical points:

• At $x=2$: $$f^{\prime \prime }(2)=12(2)-30=24-30=-6<0⟹\text{Local maximum}$$
• At $x=3$: $$f^{\prime \prime }(3)=12(3)-30=36-30=6>0⟹\text{Local minimum}$$

Step 4: Find corresponding $y$-values

• For $x=2$: $$f(2)=2(2{)}^3-15(2{)}^2+36(2)+10=16-60+72+10=38$$ So, the local maximum is at $(2,38)$.

• For $x=3$: $$f(3)=2(3{)}^3-15(3{)}^2+36(3)+10=54-135+108+10=37$$ So, the local minimum is at $(3,37)$.

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Finding the Point of Inflection

A point of inflection occurs where the concavity of the function changes, i.e., where $f^{\prime \prime }(x)=0$ or is undefined.

From Step 3: $$f^{\prime \prime }(x)=12x-30=0⟹x=\frac{30}{12}=2.5$$

Find $y$-coordinate at $x=2.5$: $$f(2.5)=2(2.5{)}^3-15(2.5{)}^2+36(2.5)+10$$ $$=2(15.625)-15(6.25)+90+10$$ $$=31.25-93.75+90+10=37.5$$

Thus, the point of inflection is at $(2.5,37.5)$.

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Part 2: Related Rates Problem (Alternative Question)

Problem Statement: A water flows into an inverted conical tank at the rate of $42\mathrm{c}{\mathrm{m}}^3/sec$. When the depth of the water is $8\mathrm{cm}$, how fast is the level rising? Assume that the height of the tank is $12\mathrm{cm}$ and the radius of the top is $6\mathrm{cm}$.

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Solution:

Step 1: Understand the geometry of the conical tank The tank is an inverted cone with:

• Height $H=12\mathrm{cm}$
• Radius at the top $R=6\mathrm{cm}$

When the water depth is $h$, the radius $r$ of the water surface can be found using similar triangles: $$\frac{r}{h}=\frac{R}{H}⟹r=\frac{R}{H}h=\frac{6}{12}h=\frac{h}{2}$$

Step 2: Express volume $V$ in terms of $h$ The volume of water in the cone is: $$V=\frac{1}{3}\pi r^{2}h=\frac{1}{3}\pi {\left(\frac{h}{2}\right)}^{2}h=\frac{\pi }{12}{h}^3$$

Step 3: Differentiate $V$ with respect to time $t$ Given that $\frac{dV}{dt}=42\mathrm{c}{\mathrm{m}}^3/sec$, we differentiate $V$ with respect to $t$: $$\frac{dV}{dt}=\frac{\pi }{12}\cdot 3h^2\cdot \frac{dh}{dt}=\frac{\pi }{4}{h}^2\frac{dh}{dt}$$

Step 4: Solve for $\frac{dh}{dt}$ when $h=8\mathrm{cm}$ Substitute $\frac{dV}{dt}=42$ and $h=8$: $$42=\frac{\pi }{4}(8{)}^2\frac{dh}{dt}$$ $$42=\frac{\pi }{4}\cdot 64\cdot \frac{dh}{dt}$$ $$42=16\pi \frac{dh}{dt}$$ $$\frac{dh}{dt}=\frac{42}{16\pi }=\frac{21}{8\pi }cm/sec$$

Final Answer: The water level is rising at a rate of $\boxed{\frac{21}{8\pi }cm/sec}$ when the depth is $8\mathrm{cm}$.

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Summary of Results (for Part 1)