Question

The maximum value of the fraction $\frac{17}{3}-{(x-\frac{4}{5})}^2$ is:

• A. $4/5$
• B. $-4/5$
• C. $17/3$
• D. $-17/3$

Answer 1

The correct option is C $17/3$

Consider a quadratic polynomial in x $f\left(x\right)=a{x}^2+bx+c$
Differentiating $f\left(x\right)$ with respect to
$\frac{df\left(x\right)}{dx}$
$=\frac{d(a{x}^2+bx+c)}{dx}$
$=2ax+b=0$ (for maxima minima)
$x=\frac{-b}{2a}$ ...(i)
Double differentiating $f\left(x\right)$ with respect to x
$\frac{d^{2}f\left(x\right)}{d{x}^2}=\frac{d^2(a{x}^2+bx+c)}{d{x}^2}$
$=2a$ ...(ii)
$\frac{d^{2}f\left(x\right)}{d{x}^2}>0$ for minimum and $\frac{d^{2}f\left(x\right)}{d{x}^2}<0$ for maximum
Therefore if 2a<0 then the polynomial has a maximum at $x=\frac{-b}{2a}$
and if 2a>0 then the polynomial has a minimum at $x=\frac{-b}{2a}$
$\frac{17}{3}-(x-\frac{4}{5}){}^2$
$=\frac{17}{3}-x^2+\frac{8x}{5}-\frac{16}{25}$
$=-x^2+\frac{8x}{5}+\frac{377}{75}$ ...(a)
Hence $a=-1,b=\frac{8x}{5},c=\frac{377}{75}$
$a<0$ hence the polynomial will have a maximum at $x=\frac{-b}{2a}$
$\frac{-b}{2a}=\frac{-8/5}{-2}=\frac{4}{5}$
Substituting in (a) we get
$-x^2+\frac{8x}{5}+\frac{377}{75}$
$=\frac{17}{3}-(x-\frac{4}{5}){}^2$
$=\frac{17}{3}-(\frac{4}{5}-\frac{4}{5}){}^2$
$=\frac{17}{3}$
Hence the maximum value of $\frac{17}{3}-(x-\frac{4}{5}){}^2$ is $\frac{17}{3}$