In order to enclose the greatest possible area with the fencing
that he has, Mr. Sanchez should plant the garden inside a circle
with a diameter of 5-ft 1.12-inches. Its area will be 20.37 square feet.
If it absolutely has to be a rectangle, then he will get the greatest
possible area when it's a square. So each side will need to be
4-ft, and the area will be 16 square feet.
I don't know how to prove either of these without calculus.
You'll just have to trust me.
Answer:
The building is 76 feet tall
Step-by-step explanation:
Since the tree is halfway beween the stake and the building, then if D is the distance between the building (at ground level) and the stake, then the tree forms with the stake a rectangle triangle with base length equal to D/2 and Height equal to 38 feets.
The stake and the top of the buidling also form a triangle rectangle with base length equal to D. Since rectangle triangle preserve proportions, if we double the length of the base without changing the angle, then the heigth should also be doubled, and as a consecuence, the height of the building is 38*2 = 76 feets.
12+25+45+65+34
= 181
Must click thanks and mark brainliest
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You can actually use either the product rule or the chain rule for this one. Observe:
• Method I:y = cos² xy = cos x · cos xDifferentiate it by applying the product rule:
![\mathsf{\dfrac{dy}{dx}=\dfrac{d}{dx}(cos\,x\cdot cos\,x)}\\\\\\ \mathsf{\dfrac{dy}{dx}=\dfrac{d}{dx}(cos\,x)\cdot cos\,x+cos\,x\cdot \dfrac{d}{dx}(cos\,x)}](https://tex.z-dn.net/?f=%5Cmathsf%7B%5Cdfrac%7Bdy%7D%7Bdx%7D%3D%5Cdfrac%7Bd%7D%7Bdx%7D%28cos%5C%2Cx%5Ccdot%20cos%5C%2Cx%29%7D%5C%5C%5C%5C%5C%5C%0A%5Cmathsf%7B%5Cdfrac%7Bdy%7D%7Bdx%7D%3D%5Cdfrac%7Bd%7D%7Bdx%7D%28cos%5C%2Cx%29%5Ccdot%20cos%5C%2Cx%2Bcos%5C%2Cx%5Ccdot%20%5Cdfrac%7Bd%7D%7Bdx%7D%28cos%5C%2Cx%29%7D)
The derivative of
cos x is
– sin x. So you have
![\mathsf{\dfrac{dy}{dx}=(-sin\,x)\cdot cos\,x+cos\,x\cdot (-sin\,x)}\\\\\\ \mathsf{\dfrac{dy}{dx}=-sin\,x\cdot cos\,x-cos\,x\cdot sin\,x}](https://tex.z-dn.net/?f=%5Cmathsf%7B%5Cdfrac%7Bdy%7D%7Bdx%7D%3D%28-sin%5C%2Cx%29%5Ccdot%20cos%5C%2Cx%2Bcos%5C%2Cx%5Ccdot%20%28-sin%5C%2Cx%29%7D%5C%5C%5C%5C%5C%5C%0A%5Cmathsf%7B%5Cdfrac%7Bdy%7D%7Bdx%7D%3D-sin%5C%2Cx%5Ccdot%20cos%5C%2Cx-cos%5C%2Cx%5Ccdot%20sin%5C%2Cx%7D)
![\therefore~~\boxed{\begin{array}{c}\mathsf{\dfrac{dy}{dx}=-2\,sin\,x\cdot cos\,x}\end{array}}\qquad\quad\checkmark](https://tex.z-dn.net/?f=%5Ctherefore~~%5Cboxed%7B%5Cbegin%7Barray%7D%7Bc%7D%5Cmathsf%7B%5Cdfrac%7Bdy%7D%7Bdx%7D%3D-2%5C%2Csin%5C%2Cx%5Ccdot%20cos%5C%2Cx%7D%5Cend%7Barray%7D%7D%5Cqquad%5Cquad%5Ccheckmark)
—————
• Method II:You can also treat
y as a composite function:
![\left\{\! \begin{array}{l} \mathsf{y=u^2}\\\\ \mathsf{u=cos\,x} \end{array} \right.](https://tex.z-dn.net/?f=%5Cleft%5C%7B%5C%21%0A%5Cbegin%7Barray%7D%7Bl%7D%0A%5Cmathsf%7By%3Du%5E2%7D%5C%5C%5C%5C%0A%5Cmathsf%7Bu%3Dcos%5C%2Cx%7D%0A%5Cend%7Barray%7D%0A%5Cright.)
and then, differentiate
y by applying the chain rule:
![\mathsf{\dfrac{dy}{dx}=\dfrac{dy}{du}\cdot \dfrac{du}{dx}}\\\\\\ \mathsf{\dfrac{dy}{dx}=\dfrac{d}{du}(u^2)\cdot \dfrac{d}{dx}(cos\,x)}](https://tex.z-dn.net/?f=%5Cmathsf%7B%5Cdfrac%7Bdy%7D%7Bdx%7D%3D%5Cdfrac%7Bdy%7D%7Bdu%7D%5Ccdot%20%5Cdfrac%7Bdu%7D%7Bdx%7D%7D%5C%5C%5C%5C%5C%5C%0A%5Cmathsf%7B%5Cdfrac%7Bdy%7D%7Bdx%7D%3D%5Cdfrac%7Bd%7D%7Bdu%7D%28u%5E2%29%5Ccdot%20%5Cdfrac%7Bd%7D%7Bdx%7D%28cos%5C%2Cx%29%7D)
For that first derivative with respect to
u, just use the power rule, then you have
![\mathsf{\dfrac{dy}{dx}=2u^{2-1}\cdot \dfrac{d}{dx}(cos\,x)}\\\\\\ \mathsf{\dfrac{dy}{dx}=2u\cdot (-sin\,x)\qquad\quad (but~~u=cos\,x)}\\\\\\ \mathsf{\dfrac{dy}{dx}=2\,cos\,x\cdot (-sin\,x)}](https://tex.z-dn.net/?f=%5Cmathsf%7B%5Cdfrac%7Bdy%7D%7Bdx%7D%3D2u%5E%7B2-1%7D%5Ccdot%20%5Cdfrac%7Bd%7D%7Bdx%7D%28cos%5C%2Cx%29%7D%5C%5C%5C%5C%5C%5C%0A%5Cmathsf%7B%5Cdfrac%7Bdy%7D%7Bdx%7D%3D2u%5Ccdot%20%28-sin%5C%2Cx%29%5Cqquad%5Cquad%20%28but~~u%3Dcos%5C%2Cx%29%7D%5C%5C%5C%5C%5C%5C%0A%5Cmathsf%7B%5Cdfrac%7Bdy%7D%7Bdx%7D%3D2%5C%2Ccos%5C%2Cx%5Ccdot%20%28-sin%5C%2Cx%29%7D)
and then you get the same answer:
![\therefore~~\boxed{\begin{array}{c}\mathsf{\dfrac{dy}{dx}=-2\,sin\,x\cdot cos\,x}\end{array}}\qquad\quad\checkmark](https://tex.z-dn.net/?f=%5Ctherefore~~%5Cboxed%7B%5Cbegin%7Barray%7D%7Bc%7D%5Cmathsf%7B%5Cdfrac%7Bdy%7D%7Bdx%7D%3D-2%5C%2Csin%5C%2Cx%5Ccdot%20cos%5C%2Cx%7D%5Cend%7Barray%7D%7D%5Cqquad%5Cquad%5Ccheckmark)
I hope this helps. =)
Tags: <em>derivative chain rule product rule composite function trigonometric trig squared cosine cos differential integral calculus</em>
Answer:
a is the answer
all work is pictured and shown