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Oduvanchick [21]
3 years ago
12

I need help, we are doing riddles in math class and Im confused.) 1. Buddy and Hermey were both born on christmas eve. When budd

y was 6 years old, hermey was half his age. If buddy turns 100 this christmas eve, how old is hermey going to be?) 2.At a holiday market, the total cost for a cup of hot cocoa and a cookie is $1.80. If a cup of hot cocoa costs $1.00 more than the cookie, how much does a cup of hot cocoa cost?) 3.
Dasher spent half as much as prancer did on holiday presents this year and cupid spent 3 times more than dasher. If the total spent between the three of them was $720, how much did they each spend on gifts?
Mathematics
1 answer:
andreev551 [17]3 years ago
8 0

Answer:

Step-by-step explanation:

for the first one the answer is 97 for the second one the it is hot coco cost 1.40$ and a cookie costs 0.40$ and for the last one it is dasher spent 120$ prancer spent 240$ and cupid spent 360$.

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<em>f(x, y)</em> = <em>x</em> ² - 4<em>xy</em> + 5

has critical points where both partial derivatives vanish:

∂<em>f</em>/∂<em>x</em> = 2<em>x</em> - 4<em>y</em> = 0   ==>   <em>x</em> = 2<em>y</em>

∂<em>f</em>/∂<em>y</em> = -4<em>x</em> = 0   ==>   <em>x</em> = 0   ==>   <em>y</em> = 0

The origin does not lie in the region <em>R</em>, so we can ignore this point.

Now check the boundaries:

• <em>x</em> = 1   ==>   <em>f</em> (1, <em>y</em>) = 6 - 4<em>y</em>

Then

max{<em>f</em> (1, <em>y</em>) | 0 ≤ <em>y</em> ≤ 2} = 6 when <em>y</em> = 0

max{<em>f</em> (1, <em>y</em>) | 0 ≤ <em>y</em> ≤ 2} = -2 when <em>y</em> = 2

• <em>x</em> = 4   ==>   <em>f</em> (4, <em>y</em>) = 12 - 16<em>y</em>

Then

max{<em>f</em> (4, <em>y</em>) | 0 ≤ <em>y</em> ≤ 2} = 12 when <em>y</em> = 0

max{<em>f</em> (4, <em>y</em>) | 0 ≤ <em>y</em> ≤ 2} = -4 when <em>y</em> = 2

• <em>y</em> = 0   ==>   <em>f</em> (<em>x</em>, 0) = <em>x</em> ² + 5

Then

max{<em>f</em> (<em>x</em>, 0) | 1 ≤ <em>x</em> ≤ 4} = 21 when <em>x</em> = 4

min{<em>f</em> (<em>x</em>, 0) | 1 ≤ <em>x</em> ≤ 4} = 6 when <em>x</em> = 1

• <em>y</em> = 2   ==>   <em>f</em> (<em>x</em>, 2) = <em>x</em> ² - 8<em>x</em> + 5 = (<em>x</em> - 4)² - 11

Then

max{<em>f</em> (<em>x</em>, 2) | 1 ≤ <em>x</em> ≤ 4} = -2 when <em>x</em> = 1

min{<em>f</em> (<em>x</em>, 2) | 1 ≤ <em>x</em> ≤ 4} = -11 when <em>x</em> = 4

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max{<em>f(x, y)</em> | 1 ≤ <em>x</em> ≤ 4, 0 ≤ <em>y</em> ≤ 2} = 21 at (<em>x</em>, <em>y</em>) = (4, 0)

min{<em>f(x, y)</em> | 1 ≤ <em>x</em> ≤ 4, 0 ≤ <em>y</em> ≤ 2} = -11 at (<em>x</em>, <em>y</em>) = (4, 2)

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