List the criteria for a 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) = 4x^3 15x^2 + 12x + 7 . OR A spherical ball of salt…

Model Answer

Part 1: Criteria for Local Maxima and Local Minima, and Finding Extrema of $f(x)=4x^3-15x^2+12x+7$

Criteria for Local Maxima and Local Minima

A function $y=f(x)$ has local maxima and local minima at a point $x=a$ if the following conditions are satisfied:

1. First Derivative Test (Necessary Condition):

   • The function must be differentiable at $x=a$.
   • The first derivative $f^{\prime }(a)=0$ (critical point).

2. Second Derivative Test (Sufficient Condition):

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

Finding Local Maxima and Minima of $f(x)=4x^3-15x^2+12x+7$

Step 1: Compute the first derivative $f^{\prime }(x)$ $$f(x)=4x^3-15x^2+12x+7$$ $$f^{\prime }(x)=\frac{d}{dx}(4x^3)-\frac{d}{dx}(15x^2)+\frac{d}{dx}(12x)+\frac{d}{dx}(7)$$ $$f^{\prime }(x)=12x^2-30x+12$$

Step 2: Find critical points by solving $f^{\prime }(x)=0$ $$12x^2-30x+12=0$$ Divide by 6: $$2x^2-5x+2=0$$ Solve using the quadratic formula: $$x=\frac{5\pm \sqrt{(-5{)}^2-4\cdot 2\cdot 2}}{2\cdot 2}$$ $$x=\frac{5\pm \sqrt{25-16}}{4}$$ $$x=\frac{5\pm 3}{4}$$ Thus, the critical points are: $$x=\frac{5+3}{4}=2\text{and}x=\frac{5-3}{4}=\frac{1}{2}$$

Step 3: Compute the second derivative $f^{\prime \prime }(x)$ $$f^{\prime \prime }(x)=\frac{d}{dx}(12x^2-30x+12)=24x-30$$

Step 4: Apply the second derivative test

• At $x=2$: $$f^{\prime \prime }(2)=24(2)-30=48-30=18>0\Rightarrow \text{Local minimum}$$ $$f(2)=4(2{)}^3-15(2{)}^2+12(2)+7=32-60+24+7=3$$ So, the local minimum is at $(2,3)$.

• At $x=\frac{1}{2}$: $$f^{\prime \prime }\left(\frac{1}{2}\right)=24\left(\frac{1}{2}\right)-30=12-30=-18<0\Rightarrow \text{Local maximum}$$ $$f\left(\frac{1}{2}\right)=4{\left(\frac{1}{2}\right)}^3-15{\left(\frac{1}{2}\right)}^2+12\left(\frac{1}{2}\right)+7$$ $$=4\left(\frac{1}{8}\right)-15\left(\frac{1}{4}\right)+6+7=\frac{1}{2}-\frac{15}{4}+13=\frac{2}{4}-\frac{15}{4}+13=-\frac{13}{4}+13=\frac{39}{4}$$ So, the local maximum is at $\left(\frac{1}{2},\frac{39}{4}\right)$.

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Part 2: Rate of Change of Radius of a Dissolving Spherical Ball

Given:

• Volume decreases at a rate of $0.75{\text{cm}}^3/\text{min}$.
• Radius $r=6\text{cm}$.

Find:

• Rate of change of radius $\frac{dr}{dt}$ when $r=6$.

Solution:

Step 1: Express volume of a sphere $$V=\frac{4}{3}\pi r^3$$

Step 2: Differentiate with respect to time $t$ $$\frac{dV}{dt}=\frac{d}{dt}\left(\frac{4}{3}\pi r^3\right)=4\pi r^2\frac{dr}{dt}$$

Step 3: Substitute known values $$\frac{dV}{dt}=-0.75{\text{cm}}^3/\text{min}(\text{negative because volume decreases})$$ $$r=6\text{cm}$$ $$-0.75=4\pi (6{)}^2\frac{dr}{dt}$$ $$-0.75=4\pi (36)\frac{dr}{dt}$$ $$-0.75=144\pi \frac{dr}{dt}$$

Step 4: Solve for $\frac{dr}{dt}$ $$\frac{dr}{dt}=\frac{-0.75}{144\pi }$$ $$\frac{dr}{dt}=\frac{-0.75}{144\times 3.1416}\approx \frac{-0.75}{452.389}\approx -0.00166\text{cm/min}$$

Final Answer: The radius of the salt ball is decreasing at a rate of $-0.00166\text{cm/min}$ (or approximately $-1.66\times 10^{-3}\text{cm/min}$) when its radius is $6\text{cm}$. The negative sign indicates that the radius is decreasing.