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March 7, 2014

# Homework Help: algebra

Posted by lee on Sunday, November 25, 2012 at 12:18pm.

Can someone help me with a couple questions and explain to me?

1)The design of a digital box camera maximizes the volume while keeping the sum of the dimensions at 6 inches if the length must be 1.5 times the height what should each dimension be? Hint let x represent one of the dimensions then define the other dimensions in term of x)

2)What are the real or imaginary solutions of the polynomial eqution?
x^4-41x^2=-400

If someone could explain these it would be helpful I don't get them

• algebra - Reiny, Sunday, November 25, 2012 at 1:50pm

1. Let the height of the box be x
then the length is 1.5x
let the width be y

so x + 1.5x + y = 6
y = 6 - 2.5x

volume = V = x(1.5x)(y) = 1.5 x^2 y
= 1.5x^2(6-2.5x)
= 9x^2 - 3.75x^3
dV/dx = 18x - 11.25x^2
= 0 for a max of V
x(18-11.25x) = 0
x = 0 --- > makes no sense (would give us minimum V)
or
x = 18/11.25 = 8/5

height = 8/5
length = 1.5(8/5) = 12/5
width = 6-2.5(8/5) = 6-4 = 2

2)
x^4 - 41x^2 + 400) = 0
(x^2 - 25)(x^2 - 16) = 0
x^2 = 25 ----> x = ± 5
or
x^2 = 16 ----> x = ± 4

• algebra - lee, Sunday, November 25, 2012 at 3:13pm

Thank you Reiny for your help.

• algebra - Bob, Friday, December 13, 2013 at 4:24am

Help... Can you give me answers for the unit test??

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