# Quiz Cal

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User Generated
Subject
Calculus
School
Grantham University
Type
Homework
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6.

----(1)
 ----------(2)
From equation 1
  



Differentiate P by x


 

For minimum perimeter


 



7.


  ----------(2)
From equation 1
  

x
y
x
y
y

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 

Differentiate P by x


 

For minimum perimeter


 




8.
  

But,   


  
Differentiate A by x



  

  
 


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6. x y Area = xy xy = 36----(1) Perimeter(P) = 2x + 2y----------(2) From equation 1 P = 2x + 2 ( P = 2x + 36 ) x 72 x Differentiate P by x 𝑑𝑃 72 =2− 2 𝑑𝑥 𝑥 For minimum perimeter dP =0 dx 2− 72 = 0; x2 x 2 = 36 𝐱 = ±𝟔 7. y y x Area = xy xy = 5000 Length of fence(L) = x + 2y----------(2) From equation 1 L = x + 2( 5000 ) x L =x+ 10000 x Differentiate P by x dL 10000 =1− dx x2 For minimum perimeter dP =0 dx 1− 10000 = 0; x2 x 2 = 10000 x = ±100 8. (𝑥, 𝑦) Area of the rectangle = xy × 2 Area of the rectangle(A) = 2xy But, y = √49 − x 2 A = 2xy A = 2x√49 − x 2 Differentiate A by x dA 1 = 2x × × (−2𝑥) + √49 − x 2 × 2 dx 2√49 − x 2 dA 2𝑥 2 = 2√49 − x 2 − dx √49 − x 2 For maximum area dA =0 dx 2√49 − x 2 − √49 − x 2 = 2𝑥 2 √49 − x 2 =0 𝑥2 √49 − x 2 𝑥 2 = 49 − 𝑥 2 2𝑥 2 = 49 𝑥=± 7 √2 y = √49 − 𝑙𝑒𝑛𝑔𝑡ℎ = 49 7 = 2 √2 14 𝑤𝑖𝑑𝑡ℎ = √2 7 √2 9. y = x4 + 6 x = −2 ∆x = dx = 0.01 using the equations to find ∆y and dy ∆𝐲 = 𝐟(𝐱 + ∆𝐱) − 𝐟(𝐱) ∆y = f(−2 + 0.01) − f(−2) ∆y = f(−1.99) − f(−2) ∆y = −1.994 + 6 − (−24 + 6) ∆y = −1.994 + 6 − (−2)4 − 6 ∆y = −0.318 f(x) = x 4 + 6 f ′ (x) = 4x 3 dy = f ′ (x)dx 𝐲 = 𝐟(𝐱) dy = f ′ (x)dx dy = 4x 3 dx dy = 4(−2)3 × 0.01 dy = −0.32 ∆𝐲 = −𝟎. 𝟑𝟏𝟖 𝐝𝐲 = −𝟎. ? ...
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