Question

Find out the maximum value of 1+8x-6x²​

Answer 1

Answer:

the Maximum value of a quadratic equation $ax^{2} + bx + c$ is given as

Step-by-step explanation:

$\frac{-D}{a}\\\\D = b^{2}-4ac\\\\D = (8)^{2}-4(1)(-6)\\\\D = 64+24\\\\D =88\\\\Now,\\\\\frac{-D}{4a} = \frac{-88}{4\times-6}\\\\= \frac{11}{3} = 3.\overline{6}$

Answer 2

GIVEN :–

• A function –

$\\ \implies\bf 1+8x-6x^{2}$

TO FIND :–

• Maximum value of given function = ?

SOLUTION :–

• Let the function –

$\\\implies\bf F(x) = 1+8x-6x^{2}$

• Differentiate with respect to 'x' –

$\\\implies\bf F'(x) = 8-(6 \times 2)x$

$\\\implies\bf F'(x) = 8-12x$

• For max/min –

$\\\implies\bf F'(x) =0$

$\\\implies\bf 8 - 12x =0$

$\\\implies\bf 8 = 12x$

$\\\implies\bf x = \dfrac{8}{12}$

$\\\implies \large{ \boxed{\bf x = \dfrac{2}{3}}}$

• Now at x = ⅔ –

$\\\implies\bf F \bigg( \dfrac{2}{3} \bigg) = 1+8 \times \dfrac{2}{3} -6\bigg( \dfrac{2}{3} \bigg)^{2}$

$\\\implies\bf F \bigg( \dfrac{2}{3} \bigg) = 1+\dfrac{16}{3} -6\bigg( \dfrac{2}{3} \bigg)\bigg( \dfrac{2}{3} \bigg)$

$\\\implies\bf F \bigg( \dfrac{2}{3} \bigg) = 1+\dfrac{16}{3} -\bigg( \dfrac{12}{3} \bigg)\bigg( \dfrac{2}{3} \bigg)$

$\\\implies\bf F \bigg( \dfrac{2}{3} \bigg) = 1+\dfrac{16}{3} -4\bigg( \dfrac{2}{3} \bigg)$

$\\\implies\bf F \bigg( \dfrac{2}{3} \bigg) = 1+\dfrac{16}{3} -\dfrac{8}{3}$

$\\\implies\bf F \bigg( \dfrac{2}{3} \bigg) = 1+\dfrac{16 - 8}{3}$

$\\\implies\bf F \bigg( \dfrac{2}{3} \bigg) = 1+\dfrac{8}{3}$

$\\\implies\bf F \bigg( \dfrac{2}{3} \bigg) = \dfrac{3 + 8}{3}$

$\\\implies \large{ \boxed{\bf F \bigg( \dfrac{2}{3} \bigg) = \dfrac{11}{3}}}$

• Hence , The maximum value is 11/3.