Question

If y = –4x^2, then the maximum value of y, is

Answer 1

Given:

A function has been provided as follows:

$y = - 4 {x}^{2}$

To find:

Maximum value of y

Concept:

We can use calculus to get the local maxima for the provided function. It would give us the maximum value of y .

It is possible by equating the 1st order differentiation of the function to zero and checking if the 2nd order differentiation is negative.

You can also use graphical representation of the function to get the maxima for that function.

Calculation:

$\therefore \: \: y = - 4 {x}^{2}$

$= > \dfrac{dy}{dx} = \dfrac{d( - 4 {x}^{2} )}{dx}$

$= > \dfrac{dy}{dx} = - 4 \times \bigg \{ \dfrac{d( {x}^{2} )}{dx} \bigg \}$

$= > \dfrac{dy}{dx} = - 4 \times \bigg \{ 2x \bigg \}$

$= > \dfrac{dy}{dx} = - 8x$

Equating the 1st order derivative to zero:

$= > \dfrac{dy}{dx} = - 8x = 0$

$= > 8x = 0$

$= > x = 0$

Now checking the 2nd order derivative:

$\therefore \: \dfrac{ {d}^{2} y}{d {x}^{2} } = \dfrac{d( - 8x)}{dx}$

$= > \: \dfrac{ {d}^{2} y}{d {x}^{2} } = - 8$

$= > \: \dfrac{ {d}^{2} y}{d {x}^{2} } < 0$

So 2nd order derivative is negative .

Hence maxima is reached at x = 0 ;

$\therefore \: \: y_{max} = - 4{x}^{2}$

$= > \: \: y_{max} = - 4 \times {(0)}^{2}$

$= > \: \: y_{max} = 0$

So maximum value for y is zero.

Final answer :

$\boxed{ \blue{ \huge{ \bold{ \: \: y_{max} = 0}}}}$

Answer 2

the maximum value of y is 0

hope it helps you.............