Thread: Max/Min Problems-$ and Diagonals

These two max min problems involve shapes with surface area, volume, etc.
The first one is asking for a ratio, not just a simple value. The second one says quadrilateral but does that mean it is a cube or something?

Originally Posted by SportfreundeKeaneKent

These two max min problems involve shapes with surface area, volume, etc.
The first one is asking for a ratio, not just a simple value. The second one says quadrilateral but does that mean it is a cube or something?

Hello,

to #13:

let x and y be the lengthes of the diagonals. Then the area of the quadrilateral is calculated by:

a = ½ * x * y

From the given conditions you know: x + y = 6 ==> y = 6 - x

Plug in the term for y into the first equation:

a(x) = ½*x*(6 - x) = -½x² + 3x

The graph of this function is a parabola opening downwards. The maximum value is at it's vertex: V(3, 4.5)

You get: x = 3, y = 3 the maximum area is 4.5 cm²

The quadrilateral is a square with diagonals which are 3 cm long.

Originally Posted by SportfreundeKeaneKent

These two max min problems involve shapes with surface area, volume, etc.
The first one is asking for a ratio, not just a simple value. The second one says quadrilateral but does that mean it is a cube or something?

Hello,

to #12:

volume of the can: V = k = π*r²*h ==> h = k/(π*r²) [1]

complete surface area: a = 2*πr² + 2*π*r*h

costs: c = 2*πr²*0.03 + 2*π*r*h*0.06 [2] Plug in the term for h from [1] into [2]:

c(r) = 2*πr²*0.03 + 2*π*r*k/(π*r²)*0.06 = 0.06*πr² + 0.12k/r

You'll get an extremum if the first derivation equals zero:

c'(r) = 0.12πr - 0.12k/r² With c'(r) = 0 you get r = ³√(k)

(It is necessary to prove if you've got a minimum. I leave this to you )

Now plug in this value into [1] and calculate h/r = 1/π