why am i getting the wrong answer? maxima and minima questions (1 Viewer)

[Lil Clutch]

ok i can do this topic ok, been working on it in the holidays but theres only two questions i can not get, i get answers but theyre completely wrong! and theyre both a similar type of question, so i was wondering if someone can help me out and see what im doing wrong. ill just put one of the questions in, its from the fitzpatrick textbook 2u- page 344, exercise 15d, question 9!

ok. heres the question, a box in the shape of a cuboid with a square base is to be made so that the sum of its dimensions is 20cm. Find the maximum volume.

its so simple! and i can do all the other maxima and minima questions-but just this one i get wrong, just to let u know of the other question similar to this which i always get wrong...its in the studymate textbook, by margaret grove, page 92, Q31. and reads: the sum of the dimensions of a box with a square base is 48m. Find the maximum volume of the box.

so u see, the two questions are similar! so why do i get them wrong?

i approach all maxima and minima questions pretty much the same, i write out an expression for what needs to be maximised or minimised, and if theres two variables, i use info given to find a relationship between the variables and obtain my expression in one variable only, then i differentiate, find stationary points, determine their nature-hence min or max and then figure it out.

so with that first question, i simply wrote the volume, which is what needs to be maximised- as V=x^2 . h (ok its hard to read this-but im trying to communicate it as best as i can-the dot is multiply! the x is sqared, for the square base, and the h is the height of the thing).

Now it sed that the sum of the dimensions=20cm

so i sed x^2 + 4xh + x^2=20, is that right? sum of the dimensions?
and that simplifies to 2x^2+4xh=20
and then rearranging the thing, i got out

h= (20-2x^2) / 4x

and then subsituted this into the volume, so that

V= x^2 . (20-2x^2) / 4x

then differentiating that, i get dv/dx = 20-6x^2 / 4 ....is that right?

well for SP's, putting dv /dx=0, i get x is + or - sqaure root 10/3....which isnt the prettiest of all numbers...but i worked with it...and checked the second derivative of it, to find that positive square root of 10/3 gave a maximum

i sub it in the volume, and get v=6.085cm^3....which is completely wrong!!

so where did i go wrong? please help...and if u dont understand ne thing i wrote-just like do the questions ureselves and explain to me how u approached it. i appraoch all the questions, practically in this similar set up, so i just think its got something to do with my derivative, or a calculation or something...help soon. thanks

 

[shafqat]

i think sum of the dimensions mean summing the lengths of the sides, ie 2x + h

 

[Slidey]

Whoa! It's maths, not english. No need to write an essay.

Sum of dimensions = 20 = 2a+b
Volume=V(x)=a^2.b
b=20-2a
V(x)=a^2(20-2a)
V'(x)=2a(20-2a)+a^2*-2=0
20a-2a^2-a^2=0
a(20-3a)=0
a=0 or a=20/3
So dimensions: 20/3 x 20/3 x 20/3

But you could have seen that coming. The maximum area for a a rectangle is given by a square. Same goes for volumes and such.

 

[dawso]

Whoa! It's maths, not english. No need to write an essay.

nice one steele, poor neggy

btw, nice signiture negs, but ill only get u the alcohol if u add me 2 it

 

[Lil Clutch]

lol. thanks people. no essay this time! but i am known for my wonderful english skills! nah, i just came online again to fix it up-coz i realised bout the dimensions thing! silly me. thanks alot anyway!

 

[dawso]

lol, renound 4 ur english aye, yeah but ur still in elliott's class so ur fuked anyway...

 

[Lil Clutch]

ok another question for my smart friends!

here goes another question ive struggled with:
studymate, margaret grove, pg 93, q45.

"from a circular disc of radius 4cm, a rectangle is cut. Find the area of the largest rectangle that can be produced"

and

q52. "show that f(x)= x^n has a minimum turning point when n is a postivie even integer.

thanks for ne help. ill go ask teachers and stuff tomorrow, but ya know- ure input is all good and now im reconsidering it, coz its nearly midnite so i guess no one will be able to help it me right now! lol. ah well. anytime! its an older topic-we did geo applications of calculus last term but i was just revising. thanks!

 

[Slidey]

"from a circular disc of radius 4cm, a rectangle is cut. Find the area of the largest rectangle that can be produced"

By inspection a square of side length 4sqrt2.

a and b are the sides on the rectangle
a^2+b^2=64 - length of diagonal of rectangle (radius 4, diameter 8, hypotenuse is diameter)
A(b)=ab
a=sqrt(64-b^2)
A(b)=bsqrt(64-b^2)
A'(b)=sqrt(64-b^2)-b^2/sqrt(64-b^2)=0
64-b^2-b^2=0
64=2b^2
b=4sqrt2

But I'm tired and I think I assumed the maximum occurs when the hypotenuse is the diameter.

 

[Pace_T]

Whoa! It's maths, not english. No need to write an essay.

vBulletin Message
You must spread some Reputation around before giving it to Slide Rule again.

 

[Lil Clutch]

vBulletin Message
You must spread some Reputation around before giving it to Slide Rule again.

what u mean??

 

[rama_v]

vBulletin Message
You must spread some Reputation around before giving it to Slide Rule again.

It means exactly what it says You cant keep repping someone...u have ppl other than slide rule too , it wont let you rep the same person over and over (for obvious reasons).

 

[Pace_T]

yeh but its funny tho, that's why i posted it
i actualy repped him ages ago if i remember correctly, so its got a long memory which is pretty good lol
...but slide rule deserves the rep lol! he's so helpful