Sunday

November 23, 2014

November 23, 2014

Posted by **sh** on Sunday, November 29, 2009 at 3:06pm.

I did Area=x^2

Area'=2x(x')

x'=area'/2xz

x'=0.8/((2)(3))

Where did I go wrong? Thanks in advance.

- Calculus -
**MathMate**, Sunday, November 29, 2009 at 3:33pmWatch the units!!!

The*length*is growing at 0.8 mm/min. So

dx/dt

=0.8 mm/min.,

=0.0008 m/min, and

x=3 m

The required answer is

d(Area)/dt=2x.dx/dt as you have worked out.

- Calculus -
**sh**, Sunday, November 29, 2009 at 3:40pmThe given rate is 0.8m/min, so the conversion is not needed. The answer given at the back is 4.8m^2/min, but the answer I have is 0.13.

- Calculus -
**MathMate**, Sunday, November 29, 2009 at 3:59pm"x'=area'/2xz

x'=0.8/((2)(3)) "

is not correct.

You have already worked out

Area'=2x(x') which is equivalent to

d(area)/dt = 2x*(dx/dt)

So substitute numerical values to get the right answer!

- Calculus -
**sh**, Sunday, November 29, 2009 at 4:10pmOh, dx/dt=0.8m/min, I thought Area' was, thanks :)

- Calculus -
**shawn**, Sunday, November 29, 2009 at 4:11pmA rectangular box with a square base and top is to be made to contain 1250 cubic feet. The material for the base costs 35 cents per square foot, for the top 15 cents per square foot, and for the sides 20 cents per square foot. Find the dimensions that will minimize the cost of the box.

- Calculus -
**MathMate**, Sunday, November 29, 2009 at 4:25pmYou would have a better chance to make a new post in case I am absent from my desk. Sometimes piggy-back questions don't get the attention that they deserve.

In any case, would you like to show me what you've done? It's a very similar problem as the previous.

You would assume the side of the square (base or top) to be x (feet).

The height can be expressed in terms of x and the volume. (V=L*W*H=x²*H)

You would then express the sum of the costs, C, made up of

1. the cost of the base = area of base * unit cost of base,

2. the cost of the sides and

3. the cost of the top.

Now differentiate C(x) with respect to x and equate C'(x)=0. Solve for x.

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