Question

The maximum value of 7 + 6x – 9x^2 is

Answer 1

Answer:

The maximum value of y is 8.

Step-by-step explanation:

We need to find the maximum value of the following expression :

$y=7+6x-9x^2$....(1)

For maximum value put $\dfrac{dy}{dx}=0$

So,

$\dfrac{d}{dx}(7+6x-9x^2)=0\\\\6-18x=0\\x=\dfrac{1}{3}$

Put x = 1/3 in equation (1) :

$y=7+6\times \dfrac{1}{3}-9\times (\dfrac{1}{3})^2\\\\y=8$

So, the maximum value of y is 8.

Answer 2

The maximum value of $7+6x-9x^{2}$ is 8.

Step-by-step explanation:

We have to find the maximum value of the following expression: $7+6x-9x^{2}$.

As we know that to find the maximum value, first we have to show that the double derivative of the expression is negative.

Let f(x) = $7+6x-9x^{2}$

Now, differentiating with respect to x, we get;

$f'(x) = 0 + 6(1) - 2 \times 9x^{(2-1)}$

$f'(x) = 6 - 18x$ --------------- [equation 1]

Now, put the value of this function equal to 0, and find the value of x;

$f'(x) = 0$

$6-18x= 0$

$18x=6$

$x = \frac{6}{18}=\frac{1}{3}$

Now, again differentiating equation 1 with respect to x, we get;

$f''(x) = 0 - 18(1)$ = -18

Since, f''(x) < 0, this means that this function will have maximum value at $x=\frac{1}{3}$.

So, the maximum value = $f(\frac{1}{3})$

      $f(\frac{1}{3}) = 7 + 6(\frac{1}{3})-9(\frac{1}{3})^{2}$

      $f(\frac{1}{3}) = 7 + \frac{6}{3}-9(\frac{1}{9})$

      $f(\frac{1}{3}) = 7 + 2-1$

      $f(\frac{1}{3}) = 8$

Hence, the maximum value of $7+6x-9x^{2}$ is 8.