A land owner is planning to build a fenced-in, rectangular patio behind his garage, using his garage as one of the "walls." He
1 answer:
Answer:
Maximum area = 800 square feet.
Step-by-step explanation:
In the figure attached,
Rectangle is showing width = x ft and the side towards garage is not to be fenced.
Length of the fence has been given as 80 ft.
Therefore, length of the fence = Sum of all three sides of the rectangle to be fenced
80 = x + x + y
80 = 2x + y
y = (80 - 2x)
Now area of the rectangle A = xy
Or function that represents the area of the rectangle is,
A(x) = x(80 - 2x)
A(x) = 80x - 2x²
To find the maximum area we will take the derivative of the function with respect to x and equate it to zero.
= 80 - 4x
A'(x) = 80 - 4x = 0
4x = 80
x =
x = 20
Therefore, for x = 20 ft area of the rectangular patio will be maximum.
A(20) = 80×(20) - 2×(20)²
= 1600 - 800
= 800 square feet
Maximum area of the patio is 800 square feet.
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There are two ways you can go about this: I'll explain both ways.
<span>
</span><span>Solution 1: Using logarithmic properties
</span>The first way is to use logarithmic properties.
We can take the natural logarithm to all three terms to utilise our exponents.
Hence, ln2ᵃ = ln5ᵇ = ln10ⁿ becomes:
aln2 = bln5 = nln10.
What's so neat about ln10 is that it's ln(5·2).
Using our logarithmic rule (log(ab) = log(a) + log(b),
we can rewrite it as aln2 = bln5 = n(ln2 + ln5)
Since it's equal (given to us), we can let it all equal to another variable "c".
So, c = aln2 = bln5 = n(ln2 + ln5) and the reason why we do this, is so that we may find ln2 and ln5 respectively.
c = aln2; ln2 =
c = bln5; ln5 =
Hence, c = n(ln2 + ln5) = n( + )
Factorise c outside on the right hand side.
c = cn(
+ 
)
1 = n( + )
= +
= 
and thus, n =
<span>Solution 2: Using exponent rules
</span>In this solution, we'll be taking advantage of exponents.
So, let c = 2ᵃ = 5ᵇ = 10ⁿ
Since c = 2ᵃ, 2 =![\sqrt[a]{c}](https://tex.z-dn.net/?f=%20%5Csqrt%5Ba%5D%7Bc%7D%20)
= 
Then, 5 =
and 10 =
But, 10 = 5·2, so 10 =·
∴ = ·
= +
and n =
<span>
</span><span>Solution 1: Using logarithmic properties
</span>The first way is to use logarithmic properties.
We can take the natural logarithm to all three terms to utilise our exponents.
Hence, ln2ᵃ = ln5ᵇ = ln10ⁿ becomes:
aln2 = bln5 = nln10.
What's so neat about ln10 is that it's ln(5·2).
Using our logarithmic rule (log(ab) = log(a) + log(b),
we can rewrite it as aln2 = bln5 = n(ln2 + ln5)
Since it's equal (given to us), we can let it all equal to another variable "c".
So, c = aln2 = bln5 = n(ln2 + ln5) and the reason why we do this, is so that we may find ln2 and ln5 respectively.
c = aln2; ln2 =
c = bln5; ln5 =
Hence, c = n(ln2 + ln5) = n(
Factorise c outside on the right hand side.
c = cn(
1 = n(
and thus, n =
<span>Solution 2: Using exponent rules
</span>In this solution, we'll be taking advantage of exponents.
So, let c = 2ᵃ = 5ᵇ = 10ⁿ
Since c = 2ᵃ, 2 =
Then, 5 =
and 10 =
But, 10 = 5·2, so 10 =
∴
and n =
I drew it out below (don't you just love my effort? XD)
You're gonna use the sine trig function to find the length of the shadow
sin26 / 1 = 43 / x; now you cross multiply
sin26 x = 43; divide both sides by sin26
x = 43 / sin26
x = 98 which is the shadow
Hope this helps!
