YT MAFS - 20210805

if you are here directly, go check out the original problem here:

https://www.youtube.com/watch?v=DRmS8VAYMnM

if you are from Youtube, NEWO, here I have a slightly different approach.

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first I have to set up the questions and label the parts. In this case,

h and k are constants, they are fixed and cannot be changed.

u and l are variables, they are flexible, they can be stretched,

because we are dealing with lengths,

u > 0, l > 0

we can imagine that this set up here is a crane lifting up an object,

somehow the object will slide along the crane such that the raised height (u) and the distance away from the base of the crane (l) can form a perfect rectangle that fits nicely under the crane.

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starting from tanα, we have

tanα = h/l = u/k

ul = hk

since h and k are constants, then ul is a constant as well, lets say that

ul = hk = v 1️⃣

l = v/u = hk/u

(this proves part 1, when h = 4, k = 3, then hk = 12 = ul = area of the rectangle)

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from the diagram, we can know that the area of the whole triangle is

A = (h+u)(l+k)/2

A = (hl + hk + ul + uk)/2

A = (hl + v + v + uk)/2

A = hl/2 + uk/2 + v

from 1️⃣, we can know that

A = hv/2u + (k/2)u + hk

A = (hhk/2)u^(-1) + (k/2)u + hk

A = (h²k/2)u^(-1) + (k/2)u + hk 2️⃣

so from this, we can adjust u and see how will the area of triangle A react.

from this, we can know that:

dA/du = (-1)(h²k/2)u^(-2) + k/2 + 0

dA/du = (-h²k/2)u^(-2) + k/2 3️⃣

we can check if the equation "A = (h²k/2)u^(-1) + (k/2)u + hk" has a minimum value by checking it's second deravitive:

d²A/du² = (-2)(-h²k/2)u^(-3)

d²A/du² = (h²k/)u^(-3)

because u > 0, so d²A/du² > 0, which means the function(/equation) A has a minimum value,

when u is something such that A is the minimum value it can have, dA/du = 0

dA/du = (-h²k/2)u^(-2) + k/2 = 0 3️⃣

(-h²)u^(-2) + 1 = 0

1 = (h²)u^(-2)

1 = h²/u²

1 = h/u

u = h

so only when u = h, then A has the minimum value it has.

If we substitute "u = h" into "A = (h²k/2)u^(-1) + (k/2)u + hk", then

Amin = (h²k/2)u^(-1) + (k/2)u + hk 2️⃣

Amin = (h²k/2)h^(-1) + (k/2)h + hk

Amin = (hk/2) + (k/2)h + hk

if we plugin h = 4(units), k = 3(units) (given from the question), then

Amin = (hk/2) + (k/2)h + hk

Amin = ((4)(3)/2) + ((3)/2)(4) + (4)(3) = 12/2 + 12/2 + 12 = 6 + 6 + 12 = 24(units²)

(this proves part 2)

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and there's a graph of A = (h²k/2)u^(-1) + (k/2)u + hk

when h = 4 and k = 3