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Question: A straight line intercept the y axis at $x=0$ and and $x$ axis at $y =0$. The line pass through a point $(3,2)$. The line show a negative slope on the graph. A triangle is form under the graph. Is there exist any minimum or maximum area of the triangle form by the line. if yes, state your reason.

Additional detail: I'm a little bit confuse. Am i rite to say that the interception coordinate at $y$-axis is $(0,y)$ and at the $x$ axis is $(x,0)$. So the line pass through 3 point that is $(0,y)$ , $(3,2)$ , and $(x,0)$. Thus, it form a triangle on the graph.

I have seen a video on youtube solving quite a similar question. but the video doesn't show me the way on how to know that the area is minimum or maximum.

Could you please show me how to solve this optimization problem using calculus. Could u show me the right way (step by step) to solve the question. What is the condition for the triangle to have the minimum or maximum area and how to write this in mathematical expression.

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If the line has to pass through $(0,b)$, $(3,2)$ and $(a,0)$ it has to satisfy the following equation: $$ y-0=\dfrac{2-0}{3-a}(x-a) $$

Observe that when $x=0$, $y=-\dfrac{2}{3-a}a$, that is $b=-\dfrac{2}{3-a}a$

Now if both $a,b$ are non-negative, then certainly the line will have negative slope (why?). Now the area $A(a,b)$ of the triangle formed by the $x$-axis, the $y$-axis and the line, is given by $A(a,b)=\dfrac{1}{2}ab$. By replacing the value of $b$ in this equation we obtain a formula for the area of the triangle as a function on $a$, $A(a)=-\dfrac{1}{2}a\dfrac{2}{3-a}a=\dfrac{a^2}{a-3}$. Now if you want to know for what value of $a$ this function has a maximum or a minimum, differential calculus is of great help.

$$ A'(a)=\dfrac{2a(a-3)-a^2}{(a-3)^2}=-\dfrac{a^2-6a}{(a-3)^2} $$

To find the critical points of this function we have to find the zeroes of $A'(a)$, that is, we need to find the solutions of the following equation: $$ -\dfrac{a^2-6a}{(a-3)^2}=0 $$ In this case the zeroes will come from the zeroes of the numerator that is $a^2-6a=0$, but those are $a=0,\quad a=6$, We have to reject $a=0$ since it does not produces a triangle. The answer is $a=6$. So the value of the area will be $A(6)=\dfrac{1}{2}\dfrac{6^2}{(6-3)^2}=2$. In order to find out whether this is a maximum or a minimum you could use the second derivative criteria: $$ A''(a)=-\dfrac{4(a-3)^3(a^2-6a)-(a-3)^4(2a-6)}{(a-3)^4}=-\dfrac{2(a-3)^3(2a^2-12a-(a-3)(a-3)}{(a-3)^4}=-\dfrac{2a^2-12a-a^2+6a-9}{(a-3)^3}=-\dfrac{a^2-6a-9}{(a-3)^3} $$

So $A''(6)=-\dfrac{6^2-6(6)-9}{(6-3)^3}=\dfrac{9}{27}=\dfrac{1}{3}>0$, the area is a minimum.

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thank you for the answer. – dgnamu Apr 12 '13 at 17:34

Let $a$ be the slope factor ($a<0$). The curve can be parameterized with the equation $y=ax+b$. Hence we know that $2=3a+b$. You can verify that the curve therefore passes through points $(0,2-3a)$ and $\left(3-\frac{2}{a},0\right)$.

The area above the x-axis, on the right of the y-axis and below the curve will be $$\frac{\left(2-3a\right)\left(3-\frac{2}{a}\right)}{2} = \frac{6-\dfrac4a-9a+6}{2}=\frac{-9a^2+12a-4}{2a}$$ (we have a right triangle).

When $a$ tends to $0^-$ or $-\infty$ this rational fraction tends to $+\infty$. So your area is not bounded.

The derivative of the area is $$\frac{(2a)(-18a+12)-2(-9a^2+12a-4)}{4a^2} $$ which has the sign of $-18a^2+8$. It reaches 0 when $a^2=\frac49$ so $a=\frac23$.

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can you show me how you get this coordinate? (0,2-3a) and (3-2/a,0) and how you get (2-3a + 3-2/a) / 2. – dgnamu Apr 12 '13 at 17:53
You have $b=2-3a$. When $x=0$, given $y=ax+b$, you must have $y=b$, so $(0,2-3a)$ is on the curve. When $y=0$, it means $ax+b=0$, so $ax+2-3a$ = 0, so $x=3-\dfrac{2}{a}$. – Benoit Apr 12 '13 at 20:04
how you get (2-3a + 3-2/a) / 2. is this a formula? am i rite to say that you are adding (y+x) / 2 to get the value above. if yes, why we adding it. is it a formula to determine whether it is a right or left triangle? if i'm not mistaken the formula for triangle is 1/2 x base x height. so form the coordinate achieve, it should be 1/2 x (3 - 2/a)(2 - 3a). – dgnamu Apr 14 '13 at 15:58
@dgnamu, you're right, my mistake, I added instead of multiplying… – Benoit Apr 15 '13 at 5:25
can u check again. (2-3a)(3-2/a) / 2 will be (6 - 4/a - 9a + 6) / 2 = (-9a - 4/a + 12) / 2 and if it is multiplying with a/a. it should be (-9a2 +12a - 4) / 2a. am i rite? – dgnamu Apr 15 '13 at 6:34

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