Find the minimum value!

I decided to expand the polynomial because it seemed there were expressions in the form of: $(a+b)(a-b)= a^{2}-b^{2}$
$(x-16)(x+16)(x-14)(x+14)= (x^{2}-256)(x^{2}-196)= x^{4} - 452 x^{2} +50176$ . In order to find the critical points (local mins, local maxes), one should take the derivative of the function above and set it to 0. This leaves:
$4x^{3}-226x=0$
$4x(x^{2}-226)=0$
$x=0$ or $x^{2} = 226$
There is no need to plug in $x=0$ because it gives the value $16 \times 14 \times 16 \times 14= somethingbig$ and we know the answer must be negative.
Notice how I kept the second possible solution in the form of $x^{2} = 226$ . This was because $x^{2}=226$ plugs in nicely to the original equation when it's in the form of: $(x^{2}-256)(x^{2}-196)$
We find that: $-A = (226-256)(226-196) = -30 \times 30$ .
$-A = -900$ So the final answer is:
$\boxed{A=900}$ .

Note that

$(x-16)(x-14)(x+14)(x+16) = \left( x^2 - 256 \right) \cdot \left( x^2 - 196 \right).$

Let this function be $f(x)$ . Then:

$f'(x) = \left( x^2 - 256 \right) \cdot \left( 2x \right) + \left( x^2 - 196 \right) \cdot \left( 2x \right) = 2x \cdot \left( 2x^2 - 452 \right).$

Note that $f'(x) = 0$ when $x = 0$ or $x = \pm \sqrt{226}$ . At $x = 0$ , it is not an absolute minimum. At $x = \pm \sqrt {226}$ , the function reaches the minimum of $-900$ , so $A = \boxed {900}$ .

First notice that $(x-16)(x+16)(x-14)(x+14) = (x^2-16^2)(x^2-14^2)=(t-16^2)(t-14^2),\quad t\ge0$ We then have a parabola (restricted to lie in the positive region of the $t$ -axis. The minimum is reached in the vertex (which lies at $t = (16^2+14^2)/2 \ge 0$ . We find $\min = \frac{-\Delta}{4} = \frac{(16^2+14^2)^2-4\cdot 16^2\cdot 14^2}{4} = \frac{(16^2-14^2)^2}{4} = \frac{(30\cdot 2)^2}{4} = \boxed{900}.$

There are different ways of approaching this problem. This is my way of approaching this problem:

It is obvious that

$(x-16)(x-14)(x+14)(x+16)$

$=(x-16)(x+16)(x-14)(x+14)$

Using the algebraic identity $a^{2}-b^{2}=(a-b)(a+b)$ , we have

$(x-16)(x+16)(x-14)(x+14)$

$=(x^{2}-16^{2})(x^{2}-14^{2})$

$=(x^{2}-256)(x^{2}-196)$

Now, since $256=226+30$ and $196=226-30$ , we have

$(x^{2}-256)(x^{2}-196)$

$=\lbrack{x^{2}-(226+30)}\rbrack\lbrack{x^{2}-(226-30)}\rbrack$

$=\lbrack{(x^{2}-226)-30}\rbrack\lbrack{(x^{2}-226)+30}\rbrack$

Now, using the algebraic identity $a^{2}-b^{2}=(a-b)(a+b)$ once again, we have

$=\lbrack{(x^{2}-226)-30}\rbrack\lbrack{(x^{2}-226)+30}\rbrack$

$=(x^{2}-226)^{2}-30^{2}$

$=(x^{2}-226)^{2}-900$

The minimum value of this expression is simply $-900$ , when $(x^{2}-226)^{2}=0$ .

Hence, $A=-(-900)=\boxed{900}$ , and we are done.

I'm a newbie here, so do advise me if I can improve on my solution in any way. And I'm not the Victor Loh who posted the question! I'm a different guy :)

This solution uses a little bit of calculus.

(x-16) (x-14) (x+14) (x+16)

can be rearranged as

(x-16) (x+16) (x-14) (x+14).

Since

a^{2}-b^{2} = (a+b) (a-b),

then

(x-16) (x+16) (x-14) (x+14)

can be expressed as

(x^{2}-16^{2}) (x^{2}-14^{2})

which is equivalent to

(x^{2}-256) (x^{2}-196).

Expanding, we obtain

x^{4}-196x^{2}-256x^{2}+256 \times 196

which is equivalent to

x^{4}-452x^{2}+256 \times 196

Let y = x^{4}-452x^{2}+256 \times 196.

According to the Power Rule, if

y = x^{n},

then

dy/dx = nx^{(n-1)}.

Hence dy/dx

= 4x^{3}-(452 \times 2)x

= 4x^{3}-904x

In order for dy/dx to be minimum, dy/dx = 0.

Hence,

4x^{3}-904{x} = 0

Dividing both sides by x, we obtain

4x^{2}-904 = 0

4x^{2} = 904

x^{2}

= 904/4

= 226

Then -a

= (x^{2}-256) (x^{2}-196)

= (226-256) (226-196)

= (-30) (30)

= -900

Hence, a = 900.

$f(x) = (x-16)(x-14)(x+14)(x+16) = (x^{2}-16^{2})(x^{2}-14^{2})$

Let $y = x^{2}$ . Then:

$f(y) = (y-16^{2})(y-14^{2}) = y^{2} - (16^{2}+14^{2}) y + 14^{2} \times 16^{2}$ .

For minimum values of f(y), $f'(y) = 0$ . So:

$2y - (16^{2} + 14^{2}) = 0$

$2y = 16^{2} + 14^{2}$

$y = 226$

$x^{2} = 226$

Let's plug this back into f(x):

$f(x) = (226-16^{2})(226-14^{2}) = -30 \times 30 = -900$

Thus, $A = 900$

-A=(x^2-256)(x^2-196) -> A=(256-x^2)(x^2-196) so dA/dx= - 2x(x^2 - 196)+ 2x(256 - x^2)= - 2x(2x^2 - 452)=0 so x^2 ={0 ، 226} -> so d^2A/dx^2 >0 when x^2=226 & <0 when x=0 so when x^2=226 the minimum value. so A=(226 - 256)(196 - 226)=900 #

On résout l'equation $f'(x)=0$ si l'on voudrait trouver le minimum de $f(x)$ . $\begin{array}{cl} &\frac d{dx}[(x-16)(x-14)(x+14)(x+16)]\\ =&\frac d{dx}[(x^2-196)(x^2-256)]\\ =&\frac d{dx}[x^4-452x^2+50176]\\ =&4x^3-904x\\ \end{array}$

$\begin{array}{rcl} f'(x)&=&0\\ 4x^3-904x&=&0\\ x^3-226x&=&0\\ x(x-\sqrt{226})(x+\sqrt{226})&=&0\\ x=-\sqrt{226}\mbox{ ou}&x=0&\mbox{ou }x=\sqrt{226} \end{array}$

• Quand $x=-\sqrt{226}$ , $f(x)=-900$ .
• Quand $x=0$ , $f(x)=50176$ .
• Quand $x=\sqrt{226}$ , $f(x)=-900$ .

Donc, le minimum de $f(x)$ est $-900$ .