Question

Find the maximum or minimum value of the quadratic expression $2x-7-5{{x}^{2}}$.

Answer 1

Hint: We know that the minimum or maximum value of a quadratic expression $y=a{{x}^{2}}+bx+c$ is $\left( \dfrac{4ac-{{b}^{2}}}{4a} \right)$ at $x=\dfrac{-b}{2a}$. If $a <0$, then the quadratic expression will have a maximum value. We will compare $2x-7-5{{x}^{2}}$ with $a{{x}^{2}}+bx+c$. Now, we will get the values of a, b and c. With these values of a, b and c, we will find the minimum or maximum values of $2x-7-5{{x}^{2}}$.

Complete step-by-step solution -
Before solving the question, we should whether a quadratic expression will have maximum value (or) minimum value.
For a quadratic expression $y=a{{x}^{2}}+bx+c$, if $a< 0$ then the quadratic expression will have maximum value. The maximum value of $y=a{{x}^{2}}+bx+c$ obtains at $x=\dfrac{-b}{2a}$. The maximum value of quadratic expression is $\left( \dfrac{4ac-{{b}^{2}}}{4a} \right)$.
In the similar way, if $a >0$ then the quadratic expression will have minimum value. The minimum value of $y=a{{x}^{2}}+bx+c$ obtains at $x=\dfrac{-b}{2a}$. The minimum value of quadratic expression is $\left( \dfrac{4ac-{{b}^{2}}}{4a} \right)$.
The given expression in this question is $2x-7-5{{x}^{2}}$.
Let us assume $y=2x-7-5{{x}^{2}}$
By rewriting the quadratic expression,
$y=-5{{x}^{2}}+2x-7$
Now we should compare $y=-5{{x}^{2}}+2x-7$ with $y=a{{x}^{2}}+bx+c$.

& a=-5.....(1) \\\ & b=2........(2) \\\ & c=-7......(3) \\\ \end{aligned}$$ We know that if $a <0$, then the quadratic expression will have a maximum value. From equation (1), it is clear that the value of a for $$y=-5{{x}^{2}}+2x-7$$ is less than zero. So, the quadratic expression $$-5{{x}^{2}}+2x-7$$ will have a maximum value.  We know that the minimum value for a quadratic expression will obtain at $$x=\dfrac{-b}{2a}$$. From equation (2) and equation (3), the maximum value of quadratic expression will obtain at $$x=\dfrac{-2}{2(-5)}=\dfrac{-2}{-10}=\dfrac{2}{10}=\dfrac{1}{5}.....(5)$$. We know that the maximum value of quadratic expression is $$\left( \dfrac{4ac-{{b}^{2}}}{4a} \right)$$. So, the maximum value of quadratic expression is $$\dfrac{4ac-{{b}^{2}}}{4a}=\dfrac{4(-5)(-7)-{{(2)}^{2}}}{4(-5)}=\dfrac{4(35)-4}{-20}=\dfrac{140-4}{-20}=\dfrac{136}{-20}=\dfrac{-68}{10}=\dfrac{-34}{5}$$. **Hence, the maximum value of $$2x-7-5{{x}^{2}}$$ is $$\dfrac{-34}{5}$$.** Note: There is an alternative method to solve this problem. A function f(x) is said to have a maximum or minimum value at the value of x where $f'(x) =0$. The value of x where f`(x)=0 is said to have a maximum value if $f”(x) <0$. The value of x where f`(x)=0 is said to have a minimum value if $f”(x) >0$. Let us assume $$f(x)=2x-7-5{{x}^{2}}......(1)$$ . $$\Rightarrow f’(x)=\dfrac{d}{dx}(2x-7-5{{x}^{2}})=2-10x$$ We have to find the value of x where f`(x) is equal to 0. $$\begin{aligned} & f’(x)=2-10x=0 \\\ & \Rightarrow 10x=2 \\\ & \Rightarrow x=\dfrac{1}{5}......(2) \\\ \end{aligned}$$ At $$x=\dfrac{1}{5}$$, $$f(x)=2x-7-5{{x}^{2}}$$ will have a maximum (or) minimum value. $$f”(x)=\dfrac{{{d}^{2}}}{d{{x}^{2}}}(2x-7-5{{x}^{2}})=\dfrac{d}{dx}\dfrac{d}{dx}(2x-7-5{{x}^{2}})=\dfrac{d}{dx}(2-10x)=-10$$ As $f''(x) <0$, so f(x) will have maximum value at $$x=\dfrac{1}{5}$$. So, substitute equation (2) in equation (1). $$f(x)=2\left( \dfrac{1}{5} \right)-7-5{{\left( \dfrac{1}{5} \right)}^{2}}=\dfrac{-34}{5}$$. Hence, the maximum value of $$2x-7-5{{x}^{2}}$$ is equal to $$\dfrac{-34}{5}$$.