# How do you determine the values of the coefficients a and b that will give a minimum value of the area, S, of the region enclosed by the curve y=-x^2+ax+b (which passes through point (1,2)) and the curve y=(1/2)x^2?

Dec 11, 2016

$S = 2$

#### Explanation:

Calling

${f}_{1} \left(x\right) = - {x}^{2} + a x + b$
${f}_{2} \left(x\right) = \frac{1}{2} {x}^{2}$

the intersection points between ${f}_{1}$ and ${f}_{2}$ are obtained solving

$\left\{\begin{matrix}{f}_{1} \left(1\right) = 2 \to - 1 + a + b = 2 \\ {f}_{1} \left(x\right) = y \to - {x}^{2} + a x + b = y \\ {f}_{2} \left(x\right) = y \to \frac{1}{x} ^ 2 = y\end{matrix}\right.$

Solving for $x , y , a$ we obtain

((x = 1/3 (3 - b - sqrt[9 + b^2]), y = 1/9 (9 - 3 sqrt[9 + b^2] + b (-3 + b + sqrt[9 + b^2]))), (x = 1/3 (3 - b + sqrt[9 + b^2]), y = 1/9 (b^2 + 3 (3 + sqrt[9 + b^2]) - b (3 + sqrt[9 + b^2]))))

with $a = 3 - b$

so the intersection points are at

${x}_{i} = \frac{1}{3} \left(3 - b - \sqrt{9 + {b}^{2}}\right)$
${x}_{s} = \frac{1}{3} \left(3 - b + \sqrt{9 + {b}^{2}}\right)$

The enclosed area is

$S \left(b\right) = {\int}_{x = {x}_{i}}^{x = {x}_{s}} \left({f}_{1} \left(x\right) - {f}_{2} \left(x\right)\right) \mathrm{dx} = \frac{1}{2} \left({x}_{i} - {x}_{s}\right) \left(- 2 b + {x}_{i}^{2} + {x}_{i} {x}_{s} + {x}_{s}^{2} - \left(3 - b\right) \left({x}_{i} + {x}_{s}\right)\right)$

Substituting now ${x}_{i}$ and ${x}_{s}$ we obtain

$S \left(b\right) = \frac{2}{27} {\left(9 + {b}^{2}\right)}^{\frac{3}{2}}$ with a minimum at $b = 0$

So finally $a = 3 , b = 0$ and $S = 2$

Attached a plot showing ${f}_{1} \left(x\right)$ blue and ${f}_{2} \left(x\right)$ red.