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.
Best Answer
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.