# Math Help - Calculus

1. ## Calculus

1. How close does the semicircle x^2 + y^2 = 16 come close to the point (3^1/2, 1)? (Hint: if you minimize the square of the distance, you can avoid dealing with square roots.)

2. The U.S. Postal Service wil accept a box for domestic shipment only if the some of its length and girth (girth being total distance around the package once) does not exceed 92 inches. What dimensions will give a box with a square end the largest possible volume?

3. The Strength of a rectangular wooden beam is proportional to its width times the square of its depth. Find the dimensions of the strongest rectangular beam that can be cut from a 12-inch diameter log.

I am in the first semester of calculus in college, so this work would have to be done with that, can anyone help me? I don't understand this at all.

2. Originally Posted by Prophet
1. How close does the semicircle x^2 + y^2 = 16 come close to the point (3^1/2, 1)? (Hint: if you minimize the square of the distance, you can avoid dealing with square roots.)
you want to minimize the distance between the points $\left( x, \pm \sqrt{16 - x^2} \right)$ and $( \sqrt{3},1)$

use the distance formula, then square both sides. minimize the resulting function

3. Originally Posted by Prophet
2. The U.S. Postal Service wil accept a box for domestic shipment only if the some of its length and girth (girth being total distance around the package once) does not exceed 92 inches. What dimensions will give a box with a square end the largest possible volume?
see here

4. #3.

We are told the strength is proportional to the product of the width and the square of the height(depth).

So, we can write $S=kwh^{2}$

S is the strength and k is a constant that depends on the wood type, etc.

We find a function that relates w and h. Draw a line from the center of the log to the corner of the beam.

We get a right triangle. $(\frac{w}{2})^{2}+(\frac{h}{2})^{2}=36$

$w^{2}+h^{2}=144$

$h^{2}=144-w^{2}$

Now, sub into the strength equation:

$S=kw(144-w^{2})=k(144w-w^{3})$

Now, differentiate, set to 0 and solve for w. h will follow.

5. Originally Posted by Prophet
3. The Strength of a rectangular wooden beam is proportional to its width times the square of its depth. Find the dimensions of the strongest rectangular beam that can be cut from a 12-inch diameter log.
Let $S$ be the strength of the beam
Let $w$ be the width of the beam
Let $d$ be the depth of the beam
Let $k$ be the proportionality constant

Since the strength is proportional to the width times the square of the depth, we have that:

$S = kwd^2$

thus, to maximize the strength, we need to maximize the dimensions of the rectangle.

now, imagine the end of the log as a circle centered at the origin. the equation of this circle is therefore given by $w^2 + d^2 = 36$ ....(i use w on the x-axis and d on the y-axis)

we want to maximize the rectangle we can get in the upper half of the circle. draw a diagram and you will realize that the length of the rectangle is given by $2w$ and the height (or depth) is given by $d$

thus we have the Area = height*width being $A = 2wd$

let's maximize this. our constraint is that $d = \sqrt{36 - w^2}$

thus we have $A = 2w \left( \sqrt{36 - w^2} \right)$ now find the values for $w$ and $d$ that maximizes this. (find $A'$ and set it to zero and solve for $w$, then plug the value you got for $w$ into the original equation to solve for $d$)

(alternatively, which may be more appropriate for this question, you can maximize the strength directly by plugging in $d = \sqrt{36 - w^2}$ into the equation for $S$. yes, i think that would be better)

6. Originally Posted by galactus
#3.

We are told the strength is proportional to the product of the width and the square of the height(depth).

So, we can write $S=kwh^{2}$

S is the strength and k is a constant that depends on the wood type, etc.

We find a function that relates w and h. Draw a line from the center of the log to the corner of the beam.

We get a right triangle. $(\frac{w}{2})^{2}+(\frac{h}{2})^{2}=36$

$w^{2}+h^{2}=144$

$h^{2}=144-w^{2}$

Now, sub into the strength equation:

$S=kw(144-w^{2})=k(144w-w^{3})$

Now, differentiate, set to 0 and solve for w. h will follow.
i like your right-triangle solution. it seems we would end up with different solutions though. i figure mine is incorrect, since i went through a lot more stuff to get to the meat of the matter. i likely made an error somewhere along the long journey

7. Originally Posted by Jhevon
you want to minimize the distance between the points $\left( x, \pm \sqrt{16 - x^2} \right)$ and $( \sqrt{3},1)$

use the distance formula, then square both sides. minimize the resulting function
Haha :P How do you minimize the function?

8. Originally Posted by Prophet
Haha :P How do you minimize the function?
find it's derivative and set it equal to zero and solve for the variable

you remember the distance formula right?

9. Originally Posted by Jhevon
find it's derivative and set it equal to zero and solve for the variable

you remember the distance formula right?
Yeah thanks.

10. So I plugged it in the formula... and then I squared both sides... but if I take the derivative, how do I work with the left side... the F^2 ???

11. Originally Posted by Prophet
So I plugged it in the formula... and then I squared both sides... but if I take the derivative, how do I work with the left side... the F^2 ???
just call it something else. so if we call D the distance formula, then the formula you want is that of D^2. but let this be a variable unto itself, that is, call D^2 a new variable, say F. so the right of your equation is now F only.

12. Originally Posted by Jhevon
just call it something else. so if we call D the distance formula, then the formula you want is that of D^2. but let this be a variable unto itself, that is, call D^2 a new variable, say F. so the right of your equation is now F only.
So with that information taken in... I have F^2 = D and so
D' = 2~3^(1/2) - 2x - 2x(16-x^(2))^(-1/2) is this right??? then I set it equal to zero?

13. Originally Posted by Prophet
So with that information taken in... I have F^2 = D and so
D' = 2~3^(1/2) - 2x - 2x(16-x^(2))^(-1/2) is this right??? then I set it equal to zero?
Seems a bit complicated :P.

14. Originally Posted by Prophet
So with that information taken in... I have F^2 = D and so
D' = 2~3^(1/2) - 2x - 2x(16-x^(2))^(-1/2) is this right??? then I set it equal to zero?
i'm not sure what you are doing here. to be clear, we want to minimize the function

$D = (\sqrt{3} - x)^2 + \left(1 - \sqrt{16 - x^2} \right)^2$

15. Originally Posted by Jhevon
i'm not sure what you are doing here. to be clear, we want to minimize the function

$D = (\sqrt{3} - x)^2 + \left(1 - \sqrt{16 - x^2} \right)^2$

So, then you take the derivative of this function and then set it equal to 0, I took the derivative and got that function... but I was just wondering if my differentiation was correct.

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