Maximizing Parallelogram Area: A Mathematical Approach

  • Thread starter Thread starter travishillier
  • Start date Start date
travishillier
Messages
15
Reaction score
0
Need some serious math help ...

Heres teh question ...

A Parallelogram has a base length of 12cm. In order to increse teh area of teh paralleogram by 54cm(squared), the length of the base is increased by 2cm and the height is incresed by 3cm . Find the area of the original parallelogram.

The solution requires me to use fully defined variables, formula(s), all steps shown using good math form and concluding statements with appropriate units .

Any help is greatly appreaciated ... PLMK what ui can do to help me ...
 
Last edited:
Mathematics news on Phys.org
area of a parallelogram = b*h

area of "old" p-gram = b*h = 12h cm
area of "new" p-gram = (b+2 cm)(h+3 cm) = (12 cm + 2 cm)(h+3 cm) = (14 cm)(h+3 cm) = 14h cm + 42 cm^2

So, the area increased by 54 cm^2, so:

(area "new" p-gram) - (area "old" p-gram) = 54cm^2

14h cm + 42 cm^2 - 12h cm = 54cm^2
2h cm + 42 cm^2 = 54cm^2
2h cm = 12 cm^2
h = 6 cm

So, the original height is 6 cm.

Original formula: A=b*h=(12 cm)(6 cm) = 72 cm^2

And, just to check with the new area...

A=(b+2)(h+3)=(12 cm +2 cm)(6 cm + 3 cm)=(14 cm)(9 cm) = 126 cm^2

And make sure the difference is 54: 126 cm^2 - 72 cm^2 = 54cm^2
 
hey, thanks a lot for teh info on this question .. you could not of helped me any more ... thanks a mill
 
Insights auto threads is broken atm, so I'm manually creating these for new Insight articles. In Dirac’s Principles of Quantum Mechanics published in 1930 he introduced a “convenient notation” he referred to as a “delta function” which he treated as a continuum analog to the discrete Kronecker delta. The Kronecker delta is simply the indexed components of the identity operator in matrix algebra Source: https://www.physicsforums.com/insights/what-exactly-is-diracs-delta-function/ by...
Fermat's Last Theorem has long been one of the most famous mathematical problems, and is now one of the most famous theorems. It simply states that the equation $$ a^n+b^n=c^n $$ has no solutions with positive integers if ##n>2.## It was named after Pierre de Fermat (1607-1665). The problem itself stems from the book Arithmetica by Diophantus of Alexandria. It gained popularity because Fermat noted in his copy "Cubum autem in duos cubos, aut quadratoquadratum in duos quadratoquadratos, et...
I'm interested to know whether the equation $$1 = 2 - \frac{1}{2 - \frac{1}{2 - \cdots}}$$ is true or not. It can be shown easily that if the continued fraction converges, it cannot converge to anything else than 1. It seems that if the continued fraction converges, the convergence is very slow. The apparent slowness of the convergence makes it difficult to estimate the presence of true convergence numerically. At the moment I don't know whether this converges or not.
Back
Top