Minimum value of expression $\displaystyle \sqrt{16b^4+(b-33)^2}$

Question

Finding point $P(a,b)$ on parabola $x=4y^2$ whose distance from the point $Q(0,33)$ is minimum and also find that minimum distance

What I try : Let coordinate of point $P$ be $(4b^2,b)$ because point $P(a,b)$ lies on $x=4y^2$

So we have $\displaystyle PQ=\sqrt{(4b^2)^2+(b-33)^2}\geq \frac{1}{\sqrt{2}}\bigg(4b^2+b-33\bigg)=\frac{4}{\sqrt{2}}\bigg[\bigg(b+\frac{1}{8}\bigg)^2-\frac{527}{64}\bigg]\geq -\frac{527}{4\sqrt{2}}$

i.e minimum occur when $\displaystyle b=-\frac{1}{8}$

But i did not understand why minimum value is negative.

Please have a look on that problem, Thanks

Note : Above we have used Inequality

$$\frac{a^2+b^2}{2}\geq \bigg(\frac{a+b}{2}\bigg)^2$$

Answer 1

The value of $b$ is the same if we are searching the minimum of $$P(b)=16b^4+b^2-66b+1089$$

The derivaitive, respect to $b$ is $$P'(b)=64b^3+2b-66$$ And the derivative of this is $$192b^2+2$$ which has no zeros and it is positive, meaning that $P'$ has only one root, and it is not zero, and that $P$ is convex everywhere. So $P(b)$ has only one minimum.

Since $P'(1)=0$, the minimum of $P$ is $P(1)=1040$.

Answer 2

Hint: We can directly look at when $|PQ|^2$ is minimised, using univariate calculus or: $$(4b^2)^2+(b-33)^2=(b-1)^2(16b^2+32b+49)+1040$$

Answer 3

Finding point $P(a,b)$ on parabola $x=4y^2$ whose distance from the point $Q(0,33)$ is minimum and also find that minimum distance

Let us find $b$ such that the slope of the line passing through $(4b^2,b)$ and $(0,33)$ is equal to the slope of the normal at $(4b^2,b)$.

From $x=4y^2$, we get $1=8yy'$. So, the slope of the normal at $(4b^2,b)$ is given by $-8b$.

So, we have $\frac{b-33}{4b^2-0}=-8b$, i.e. $32b^3+b-33=0$ which can be written as $(b-1)\underbrace{(32b^2+32b+33)}_{\text{positive}}=0$. So, we get $b=1$.

The equation of the tangent line at $(4,1)$ is $y=\frac x8+\frac 12$. Also, the equation of the parabola with $y\ge 0$ is $y=\frac{\sqrt x}{2}$. Now, for every $x\ge 0$, $\frac x8+\frac 12\ge \frac{\sqrt x}{2}$ holds since this is equivalent to $(\sqrt x-2)^2\ge 0$ whose equality is attained only when $x=4$.

So, the point we want is $\color{red}{(4,1)}$, and the minimum distance is $\sqrt{(4-0)^2+(1-33)^2}=\color{red}{4\sqrt{65}}$.

Answer 4

We introduce a new $\color{#c00}{\alpha}$ variable, to get the minimum of $P(b):={16b^4+(b-33)^2}$ .

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Let $\alpha>0$ . Then aplying the AM-GM inequality, we have :

$$ \begin{align}&16b^4+\alpha^2≥8b^2\alpha\\ \implies&16b^4≥8b^2\alpha-\alpha^2 \end{align} $$

This leads to the following quadratic respect to the variable $\color{#c00}{b}$ :

$$ \small {\begin{align}P(b)&≥8b^2\alpha+(b-33)^2-\alpha^2\\ &=\color{#c00}{b^2}(8\alpha+1)-66\color{#c00}{b}+(33^2-\alpha^2)\\ &=Q(b)\end{align}} $$

Then, the standard completing the square yields :

$$ \small{\begin{align}Q(b):=(8\alpha+1)\left(b-\frac {33}{8\alpha+1}\right)^2-\frac {\Delta_b}{4(8\alpha+1)} \end{align}} $$

Thus, we reduce the problem to finding the minimum of $Q(b)$, which is equivalent to determine when the equality case occurs :

$$ \begin{align}\begin{cases}\alpha=4b^2, \thinspace\thinspace\thinspace\text{by AM-GM}\\ b=\dfrac {33}{8\alpha+1}, \thinspace\thinspace\thinspace\text {by Discriminant}\end{cases}\end{align} $$

The above system of equations yields :

$$ \begin{align}0&=32b^3+b-33\\ &=33b^3-b^3+b-33\\ &=33(b^3-1)-(b^3-b)\\ &=\small{33(b-1)(b^2+b+1)-b(b-1)(b+1)}\\ &=(b-1)\underbrace{(32b^2+32b+33)}_{\Delta_b\thinspace<\thinspace 0} \end{align} $$

Finally, we obtain the equality case $b=1$, which leads to the final conclusion :

$$ \begin{align}P(b):&=16b^4+(b-33)^2\\ &≥33b^2-66b+1073\\ &=33(b-1)^2+1040\\ &≥1040\thinspace\thinspace\thinspace\thinspace\thinspace\tiny{\blacksquare}\end{align} $$

Answer 5

We seek a value $m$ for which the polynomial

$16b^4+(b-33)^2-m=16b^4+b^2-66b+1089-m$

has a doule zero (where the polynomial touches but foes not cross the horizontal axis) at some point $b=b_0$. The smallest $m$ satisfying this condition is the minimum value.

We render the factorization

$16b^4+b^2-66b+1089-m=(b-b_0)^2(16b^2+\alpha_1 b+\alpha_0)).$

Expanding right side and matching cubic, quadratic and lear terms in that order yields

$\alpha_1=32b_0$

$1=-\alpha_0-2\alpha_1b_0+16b_0^2\implies \alpha_0=2\alpha_1b_0-16b_0^2-1=48b_0^2-1$

$-2b_0\alpha_0+b_0^2\alpha_1=-66\implies \color{blue}{64b_0^3+2b_0-66=0}.$

The blue equation is the "zero derivative condition" which we derived for this polynomial minimization without calculus. Clearly $b_0=1$ is a root. Then factoring gives

$2(b_0-1)(32b_0^2+32b_0+33)=0,$

where the quadratic term has a negative discriminant and thus no more real roots. So the minimal distance must occur at $y=b_0=1,x=4y^2=4$ and we calcuate $m=1040$, thus the distance is $\sqrt{m}=4\sqrt{65}$.

Answer 6

By AM-GM $$16b^4+(b-33)^2=16b^4+(b-1-32)^2=16b^4+(b-1)^2-64(b-1)+1024\geq$$ $$\geq16b^4-64(b-1)+1024=16(b^4-4b+3)+1040\geq16\left(4\sqrt[4]{b^4\cdot1^3}-4b\right)+1040=$$ $$=64\left(|b|-b\right)+1040\geq1040.$$ The equality occurs for $b=1$ and since $f(x)=\sqrt{x}$ increases, we obtain: $$\min\sqrt{16b^4+(b-33)^2}=\sqrt{1040}.$$

Answer 7

We want to minimise $16x^4+(x-33)^2$, its derivative is $2(32x^3+x-33)$. This has a root at $x=1$ (either notice this or try with the Rational Root Theorem). Then we can factor it as $2(x-1)(32x^2+32x+33)$, and the second polynomial does not have real roots. In fact, the derivative is positive for $x>1$ and negative for $x<1$. Hence the minimum is attained at $x=1$ and it is equal to $4\sqrt{65}$.