Answer:
p = 25,20 cm
h = 34,32 cm
A(min) = 864,86 cm²
Step-by-step explanation:
Let call p and h dimensions of a poster, ( length , and height respectively) and x and y dimensions of the printed area of the poster then
p = x + 8 and h = y + 12
Printed area = A(p) = 384 cm² and A(p) = x*y ⇒ y = A(p)/x
y = 384 / x
Poster area = A(t) = ( x + 8 ) * ( y + 12 ) ⇒ A(t) = ( x + 8 ) * [( 384/x ) + 12 ]
A(t) = 384 + 12x + 3072/x + 96 A(t) = 480 + 3072/x + 12x
A(t) = [480x + 12x² + 3072 ] / x
A´(t) = [(480 + 24x )* x - (480x + 12x² + 3072]/x²
A´(t) = 0 [(480 + 24x )* x - 480x - 12x² - 3072] =0
480x + 24x² -480x -12x² - 3072 = 0
12x² = 3072 x² = 296
x = 17,20 cm and y = 384/17,20 y = 22,32 cm
Notice if you substitu the value of x = 17,20 in A(t) ; A(t) >0 so we have a minimun at that point
Then dimensions of the poster
p = 17,20 + 8 = 25,20 cm
h = 22.32 + 12 =34,32 cm
A(min) = 25,20 *34.32
A(min) = 864,86 cm²
12x=4(x-5)
distributive property to get rid of parenthesis 12x=4x-20
subtract x from one side to get the variable on one side 8x=-20
divide 8 by -20
x=-2.5
Answer:
6x -10 = 5x+12 (vertically opposite angles)
6x-5x =12+10
x= 22
Step-by-step explanation:
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Answer:
Step-by-step explanation:
Let x represents the number of nights Jack worked and y represents the number of nights Diane worked.
1. The number of nights Diane is scheduled to work is no more than four times the number of nights Jack is scheduled to play. Then
2. Diane will work at least 10 times before the concert. Then
3. Jack earns $50 per night that he plays, then he earned $50x in x nights. Diane earns $25 each night she works, then she earns $25y in y nigths. They need at least $750, so
4. We get the following set of constraints to model the problem:
