Let f(x) = 4-x^2. For 0

Question

Let f(x) = 4-x^2. For 0<p<2, let A(p) be the area of the triangle formed by the coordinate axes and the line tangent to the graph of f at the point (p,4-p^2).

a. Find A(2)

b. For what value of p is A(p) a minimum?

anonymous · Answer

i gotta draw this one and see if i get any ideas
did you draw it?

anonymous · Answer

i guess the best thing is to follow the instructions and find $A(2)$ first
did you get that one?

anonymous · Answer

lol $A(2)=0$   doh
i thought is asked for the maximum, but it asked for the minimum!

anonymous · Answer

yes so that would be 0

anonymous · Answer

oh wait, let me try again maybe it is not zero

anonymous · Answer

but wouldn't you just plug I tin the equation

anonymous · Answer

the point is on the $x$axis but the triangle exists

anonymous · Answer

if $p=2$ then the point is $(2,0)$ on the $x$ axis, but the slope is $-4$ right?

anonymous · Answer

you following?  the derivative of $4-p^2$ is $-2p$ and if $p=2$ then the slope is $-4$

anonymous · Answer

right

anonymous · Answer

|dw:1398741002491:dw|

anonymous · Answer

i got the $8$ because the slope is $-4$ and i solved$$\frac{b-0}{0-2}=-4$$ giving $b=8$

anonymous · Answer

then 
$$A(2)=\frac{1}{2}8\times 2=8$$

anonymous · Answer

ohhh right I see

anonymous · Answer

but that is just the solution for $A(2)$ now we have to come up with a function $A(x)$ to minimize

anonymous · Answer

wait sorry could you please explain the solving for b part again

anonymous · Answer

sure

anonymous · Answer

the line crosses the $y$ axis somewhere, and we need this point because it is the height of the triangle
where the line crosses the $y$ axis is at $(0,b)$ for some $b$
we already have one point $(2,0)$  so i computed the slope between the pionts 
$$(0,b),(2,0)$$ as 
$$m=\frac{b-0}{0-2}=\frac{b}{-2}$$

anonymous · Answer

since that slope has to be $-4$ we know $\frac{b}{-2}=4\implies b=8$

anonymous · Answer

oops i meant 
$$\frac{b}{-2}=-4\iff b=8$$

anonymous · Answer

i am still trying to find a formula for $A(p)$ though...

anonymous · Answer

wait but what I don't understand is that don't we know that the curve passes through (0,4) ?

anonymous · Answer

yes it does, but the tangent line will be above the curve

anonymous · Answer

take a look at this picture of the tangent line $y=-4x+8$ and the curve $y=4-x^2$

anonymous · Answer

http://www.wolframalpha.com/input/?i=y%3D4-x^2%2Cy%3D-4x%2B8 lol this picture

anonymous · Answer

Oh right right IDK why I was drawing the line inside the curve alright makes sense

anonymous · Answer

oh maybe we can find a general formula for the tangent line in terms of $p$ and then solve for where it crosses the $x$ axis and the $y$ axis

anonymous · Answer

i bet we can do that without too much trouble

anonymous · Answer

think i got it
you want to try it?

anonymous · Answer

You wouldn't draw it like this?? |dw:1398742054924:dw| like I thought a tan slope goes thru themiddle

anonymous · Answer

yes

anonymous · Answer

the picture i drew in wolfram used the point on the $x$ axis $(2,0)$ because we were given $p=2$ so the point was $(2,0)$ but in general it will look like what you drew

anonymous · Answer

ready to find the equation of the line?

anonymous · Answer

ohh okok

anonymous · Answer

we could redo this at $(1,3)$ if you like 
it shouldn't be that hard now that we did it once

anonymous · Answer

why don't we try that first?  might be clearer how to proceed

anonymous · Answer

or do you want to try it yourself? 
either way

anonymous · Answer

Cool ok, a, makes sense as for b cant we just do that on the calculator ?

anonymous · Answer

we can do it by hand, not sure a calculator will be useful, and it certainly won't be useful in finding $A(p)$

anonymous · Answer

at the point $(1,3)$ the slope is $-2$right?  and so the equation for the line is $y-3=-2(x-1)$ which we could multiply out etc but lets leave it like that

anonymous · Answer

it crosses the $y$ axis where $x=0$ so we can solve 
$$y-3=-2(0-1)$$ or just 
$$y-3=2$$ making $y=5$

anonymous · Answer

we also need to know where it crosses the $x$ axis where $y=0$ we can solve 
$$0-3=-2(x-1)$$
or 
$$-3=-2(x-1)\
\frac{3}{2}=x-1\
\frac{5}{2}=x$$

anonymous · Answer

then 
$$A(1)=\frac{1}{2}\times 5\times \frac{5}{2}$$

anonymous · Answer

how are we doing so far?

anonymous · Answer

makes sense so far

anonymous · Answer

i.e. do you understand how i got that or do you have any questions?

anonymous · Answer

now comes the hard part, we have to do it with $p$ instead of $2$ or $1$

anonymous · Answer

ready?

anonymous · Answer

yes but before you start may you give me a visual representation

anonymous · Answer

yes but before you start may you give me a visual representation

anonymous · Answer

here is the one we just did scale has changed a bit http://www.wolframalpha.com/input/?i=y%3D4-x^2%2Cy-3%3D-2%28x-1%29+

anonymous · Answer

that one is a little confusing, because they omitted the $y$ axis, should look more like this

anonymous · Answer

okay wait could you explain what we are trying to find for B then? it is the Minimum point of the curve not the tan?

anonymous · Answer

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