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**GrigOrig99** The question statement indicates that the area will be a maximum, but in my attempt you can see that I have arrived at a positive value of 6, suggesting a minimum. I think this has something to do with how I've applied the modulus when calculating the area of the triangle. Specifically, I 'probably' can't define $\displaystyle \frac{1}{2}|x^3-6x^2|$ as either $\displaystyle \frac{x^3-6x^2}{2}$ or $\displaystyle \frac{-x^3+6x^2}{2}$. Furthermore, as you can see, my final answer does not match that of the text book. However, I've been told that my answer is correct, so long as the value expressed is a maximum. Can anyone help me out?

Many thanks.

Q. The given diagram (see attachment) shows part of the graph of $\displaystyle y=x^2$. p(x, y) is a point on the curve and $\displaystyle a=(6,0)$. bp is perpendicular to ab. **(i)** Express the coordinates of p in terms of x only, **(ii)** Find the value of x, if the area of the $\displaystyle \triangle abp$ is a maximum &, hence, find the maximum area.

**Attempt:** **(i)** If $\displaystyle y=x^2$, then $\displaystyle p(x,x^2)$

**(ii)** From the diagram, we can infer that $\displaystyle b=(x,0)$.

Area of $\displaystyle \triangle abp$: $\displaystyle \frac{1}{2}|(y_2-y_1)(x_1-x_3)-(x_2-x_1)(y_1-y_3)|$

Subbing in the vertices of the triangle: $\displaystyle \frac{1}{2}|(0-0)(6-x)-(x-6)(0-x^2)|=\frac{1}{2}|0-(-x^3+6x^2)|=\frac{1}{2}|x^3-6x^2|=$

$\displaystyle \frac{x^3-6x^2}{2}$

Find derivative to determine value of x: $\displaystyle \frac{dA}{dx}=\frac{1}{2}(3x^2-12x)=0\rightarrow3x^2-12x=0\rightarrow x(x-4)=0\rightarrow x= 0$ or $\displaystyle x=4$

Second derivative determines maximum/ minimum area: $\displaystyle \frac{d^2A}{dx^2}=3x-6\rightarrow3(4)-6=6>0$, where area is a minimum when $\displaystyle x=4$

Area: $\displaystyle \frac{x^3-6x^2}{2}=\frac{4^3-6(4)^2}{2}=\frac{64-96}{2}=\frac{-32}{2}=-16$

**Ans.:** (From text book): x = 2, Area = 8