Wednesday

October 22, 2014

October 22, 2014

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

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 1:50pm1. 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 3:13pmThank you Reiny for your help.

- algebra -
**Bob**, Friday, December 13, 2013 at 4:24amHelp... Can you give me answers for the unit test??

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