# Thread: Help with pre-claculus homework would be greatly appreciated

1. ## Help with pre-claculus homework would be greatly appreciated

I've been stuck on this worksheet for days now.. we're working on derivatives now and my teacher gave us a few word problems and for most of them i think i got the correct function but am a bit doubtful when it come to my abilities (especially after my teacher told my friend behind my back that i was not reliable source )
Anyways, the first of the equations should be easy i know but i just want to check on it. Question: Find the point(s) on the graph of y= 4 - x^2 that is closest to the point(0,2)

For this one i know that we use the distance formula and the domain is all real numbers, but im not quite sure how to replace the "y" in the equation for x so that there would be only one variable.

2nd question: A rectangular page is to contain 24 square inches of print the margins at the top and at the bottom of the page are 1.5 inches each. THe margins on each side arae 1 inch. What should the dimensions of the page be so that the least amout of paper is used?

The function would be f(x) = x((24/x-2) + 3) and the derivative would be f(x)' = ((24/x-2) + 3) + x((-24/(x-2)^2)). Think the critical points are 0 and 2 not sure about the domain though.

The last question that i was not sure about is: Two posts, one 12 feet high and the other 28 feet high, stand 30 feet apart. They are to be stayed by two wires, attached to a single stake, running from the ground level to the top of each post. Where should teh stake be placed to use the least wire?

The function for this one i found out would be f(x) = sqrt(x^2 + 144) + sqrt(x^2 - 60x + 1684), but im having mad problems finding the deriviative and critical points for this one.

Any help with these would be greatly appreciated and hope the additional info helps.

Thanks

2. Originally Posted by rm1991
I've been stuck on this worksheet for days now.. we're working on derivatives now and my teacher gave us a few word problems and for most of them i think i got the correct function but am a bit doubtful when it come to my abilities (especially after my teacher told my friend behind my back that i was not reliable source )
Anyways, the first of the equations should be easy i know but i just want to check on it. Question: Find the point(s) on the graph of y= 4 - x^2 that is closest to the point(0,2)

For this one i know that we use the distance formula and the domain is all real numbers, but im not quite sure how to replace the "y" in the equation for x so that there would be only one variable.
The square of the distance between the points (0, 2) and (x, y) is
$D = d^2 = x^2 + (y - 2)^2$

If the point (x, y) is on the curve $y = 4 - x^2$ then this is
$D = x^2 + (4 - x^2 - 2)^2 = x^4 + 5x^2 + 4$

You want the minimum distance so you wish to minimize D. (The minimum of the distance and minimum of the distance squared obviously correspond to the same x value.)

So take the first derivative of D and set it to 0. What x values do you get? Which is the correct answer?

-Dan

3. Originally Posted by rm1991

The function for this one i found out would be f(x) = sqrt(x^2 + 144) + sqrt(x^2 - 60x + 1684), but im having mad problems finding the deriviative and critical points for this one.

Thanks
Ok so we have that

$f(x)=\sqrt{x^2+144}+\sqrt{x^2-60x+1684}=(x^2+144)^{\frac{1}{2}}+(x^2-60x+1684)^{\frac{1}{2}}$

Now, post back if you arent versed in the quicker methods of differentiation and are still doing the quotient difference.

But utilizing the chain rule we get

$f'(x)=\frac{1}{2}(x^2+144)^{\frac{-1}{2}}\cdot(2x)+\frac{1}{2}(x^2-60x+1684)^{\frac{-1}{2}}\cdot(2x-60)$

Simplifying a little we get

$\frac{x}{\sqrt{x^2+144}}+\frac{x-30}{\sqrt{x^2-60x+1684}}$ $=\frac{x\sqrt{x^2-60x+1684}}{\sqrt{x^2+144}\sqrt{x^2-60x+1684}}+\frac{(x-30)\sqrt{x^2+144}}{\sqrt{x^2-60x+1684}\sqrt{x^2+144}}=$ $\frac{x\sqrt{x^2-60x+1684}+(x-30)\sqrt{x^2+144}}{\sqrt{(x^2+144)(x^2-60x+1684)}}$

Now setting the numerator equal to zero we get

$x\sqrt{x^2-60x+1684}=$ $-(x-30)\sqrt{x^2+144}\Rightarrow{x^2(x^2-60x+1684)=(x-30)^2(x^2+144)}\Rightarrow{x\in\mathbb{C}}$

Now this was purely for teaching purposes, I woul suggest going back and looking at your set up, for never in precalc would that be that hard, seconldy, you cannot have an imaginary length. How exactly did you come across that, maybe if you show us we can more easily help you

Mathstud

4. yea sure thing... this is how i imagined it

unfortunately he gives us problems like this all the time.... sometimes i wonder how i manage to keep sanity through out the year ha
hope this helps