# Calculus problem help (calc 1 AB)

• Dec 30th 2007, 01:18 PM
tommyjay117
Calculus problem help (calc 1 AB)
I need some help with this Calc problem from Calc AB (calc 1)

A Tank with dimensions w x l x h has a rectangular base, rectangular sides and an open top. It is to be constructed so that its width is 4 meters and its volume is 36 cubic meters. Building the tank costs $10 per sqaure meter for the base and$5 per sqaure meter for the sides.

a) Write the cost function, C(h) in terms of h, where h is the height of the tank. What is the domain of C(h)

b)Find the critical points of C and analyze whther these critical points are the locations of extrema. Justify your answers.

c) What are the dimensions of the least expensive tank? Indicate units of measure.

d) What is the cost of the least expensive tank?

If you could help me that would be great, I've been staring at this for an hour, and for some reason I can't imagine how to do it.

Edit: Whoops, mod can move this to homework help if they like. Sorry.
• Dec 30th 2007, 01:47 PM
galactus
Since the width is 4, then the volume can be expressed as 4LH=36.

The surface area is $S=4L+2(4H)+2(LH)$

The cost would then be $C=10(4L)+5(2)(4H)+5(2)(LH)=40L+40H+10LH$

But from the volume equation, $L=\frac{36}{4H}$.

$C=40\cdot\frac{36}{4H}+40H+10\cdot\frac{36}{4H}\cd ot{H}=40H+\frac{360}{H}+90$

Now, differentiate, set to 0 and solve for H to find the height requirement which will minimize the surface area and cost. You can then find L by subbing it back into the length equation we found. We already know the width is 4.
• Dec 30th 2007, 09:19 PM
tommyjay117
Thanks, that's a big help. I always forget to substiture the variables from the other formulas that I already know. This is another I was having trouble with, any ideas?

Let f(x)= 12-x² for x ≥ 0 and f(x) ≥ 0

a) The line tangent to the graph of f at the point (k, f(k)) intercepts the x-axis at 4 = x. What is the value of k?

b) An isosceles triangle whose base is the interval from (0,0) to (c,0) has its vertex on the graph of f. For what value of c does the triangle have maximum area? Justify your answer.

I know for a that the slope of the line tangent to f is f'(x) = 2x. So the line is y = 2x(x-k) + f(k). it intercepts at (4,0) so 0 = 2(4)(4-k) + 0 => 0 = 32-8k => k = 4. Right? I don't know what to do for b though.
• Dec 30th 2007, 09:50 PM
Isomorphism
Quote:

Originally Posted by tommyjay117
Thanks, that's a big help. I always forget to substiture the variables from the other formulas that I already know. This is another I was having trouble with, any ideas?

Let f(x)= 12-x² for x ≥ 0 and f(x) ≥ 0

a) The line tangent to the graph of f at the point (k, f(k)) intercepts the x-axis at 4 = x. What is the value of k?

b) An isosceles triangle whose base is the interval from (0,0) to (c,0) has its vertex on the graph of f. For what value of c does the triangle have maximum area? Justify your answer.

I know for a that the slope of the line tangent to f is f'(x) = 2x. So the line is y = 2x(x-k) + f(k). it intercepts at (4,0) so 0 = 2(4)(4-k) + 0 => 0 = 32-8k => k = 4. Right? I don't know what to do for b though.

A point on f is of form (x,12-x^2)...
Area of a triangle is 0.5 x base x height
Base length is c and now we have to calculate height.

height = $\sqrt{(x-\frac{c}2)^2+(12 - x^2)^2}$
So maximize height :D
• Dec 31st 2007, 10:46 AM
galactus
Quote:

Let f(x)= 12-x² for x ≥ 0 and f(x) ≥ 0

a) The line tangent to the graph of f at the point (k, f(k)) intercepts the x-axis at 4 = x. What is the value of k?
You can use $y-y_{1}=m(x-x_{1})$ and solve for x.

$12-x^{2}-0 = -2x(x-4)$

$x^{2}-8x+12=(x-2)(x-6)=0$

We can disregard x=6 and pay attention to x=2

That means the line is tangent at (2,8). You can now use this point and (4,0) to find its equation.
• Jan 1st 2008, 11:13 AM
tommyjay117