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Question: The minimum value of \({x^2} - 12x + 40\) and the maximum value of \(24x - 8 - 9{x^2}\) is?...

The minimum value of x2−12x+40{x^2} - 12x + 40 and the maximum value of 24x−8−9x224x - 8 - 9{x^2} is?

Explanation

Solution

Hint-In this question we have to find the minimum and maximum value of the given quadratic equations. Check for the coefficient of highest power term to be positive or negative and apply the respective formula for maximum and minimum value of a quadratic equation.

Complete step-by-step answer:
We have to find the minimum value of x2−12x+40{x^2} - 12x + 40 and the maximum value of 24x−8−9x224x - 8 - 9{x^2}.
⇒\Rightarrow Now if we have a quadratic equation of the form ax2+bx+c=0{\text{a}}{{\text{x}}^2} + bx + c = 0such that a>0a > 0 then the minimum value of this quadratic equation is 4ac−b22a\dfrac{{4ac - {b^2}}}{{2a}}at the point x=b2a\dfrac{b}{{2a}} and in this case the maximum value doesn’t exist as the parabola formed is opening upwards and the domain of this equation is R (that is real numbers) thus the max value can go up to infinity.
⇒\Rightarrow Now if we talk about a quadratic equation of the form ax2+bx+c=0{\text{a}}{{\text{x}}^2} + bx + c = 0 such that a<0a < 0 then the maximum value of this quadratic equation is 4ac−b22a\dfrac{{4ac - {b^2}}}{{2a}} at point x=−b2a\dfrac{{ - b}}{{2a}} and in this case the minimum does not exist as the parabola formed is opening downwards and the domain of this equation is R (that is real numbers) thus the min value can go up to minus infinity.
Now in the equation x2−12x+40{x^2} - 12x + 40 a=1, b=-12 and c=40
As a>0a > 0 because 1>0, hence using the above mentioned concept it does not have a maximum value and the minimum value is given by 4ac−b22a\dfrac{{4ac - {b^2}}}{{2a}}.
So substituting the values we get minimum value = 4(1)×40−(−12)22×1=8\dfrac{{4(1) \times 40 - {{\left( { - 12} \right)}^2}}}{{2 \times 1}} = 8
Now in the equation 24x−8−9x224x - 8 - 9{x^2}, a=-9, b=24, c=-8.
As a<0a < 0 because -9<0, hence using the above mentioned concept at does not have a minimum value and the maximum value is given by 4ac−b22a\dfrac{{4ac - {b^2}}}{{2a}}.
So substituting the values we get the maximum value = 4(9)×(−8)−(24)22×9=16\dfrac{{4(9) \times ( - 8) - {{\left( {24} \right)}^2}}}{{2 \times 9}} = 16
Note – Whenever we face such types of problems the key concept is to understand whether the parabola formed by the equation given is opening up parabola or opening down parabola, it all depends upon the value of the coefficient of highest power of that equation. Then applying the concept mentioned above we can get the answer.